User:Ayutu's Integral

1.$$\int_{0}^{1}\frac{\cos(\ln x)\ln x}{1+x}\mathrm dx=\frac{1-\pi^2\color{blue}\coth(\pi)\color{red}\operatorname{csch}(\pi)}{2}$$

2.$$\int_{0}^{\pi/2}x\cos(8x)\ln\left(\frac{1+\tan x}{1-\tan x}\right)\mathrm dx=\frac{\pi}{12}$$

3.$$\int_{0}^{\pi/2}\cos(2x)\ln^2\left(\frac{1+\sin x}{1-\sin x}\right)\mathrm dx=-4\color{green}\pi$$

4.$$\int_{0}^{\pi/2}\cos(x)\ln^2\left(\frac{1+\sin x}{1-\sin x}\right)\mathrm dx=\color{red}2\zeta(2)$$

5.$$\int_{0}^{\pi/2}\cos^2(x)\ln^2\left(\frac{1+\sin x}{1-\sin x}\right)\mathrm dx=\color{red}\pi\left(\left(\color{blue}\frac{\pi}{2}\right)^2-2\right)$$

6.$$\int_{0}^{\pi/2}x\cos(x)\ln^2\left(\frac{1+\sin x}{1-\sin x}\right)\mathrm dx=\color{red}\frac{32}{9}$$

7.$$\int_{-\infty}^{+\infty}\frac{e^x+1}{(e^x+x+1)^2+\pi^2}\mathrm dx=1$$

8.$$8\int_{0}^{\infty}\frac{\ln x}{x}\left(e^{-x}-\frac{1}{\sqrt[4]{1+8x}}\right)\mathrm dx=-32C+4\gamma^2-5\pi^2$$

9.$$\int_{0}^{1}\frac{x+x^2+\cdots+x^{2n}-2nx}{(1+x)\ln x}\mathrm dx=\ln\left[\color{red}\left(\frac{2}{\pi}\right)^n(2n)!!\right]$$

10.$$\int_{0}^{1}\frac{x+x^2+x^3-3x}{(1+x)^3\ln x}\mathrm dx=\color{blue}\ln\left(\frac{\pi}{2}\right)-\frac{7}{4}\cdot \color{red}\frac{\zeta(3)}{\pi^2}$$

11.$$\int_{0}^{1}\frac{x+x^2+\cdots+x^5-5x}{(1+x)^3\ln x}\mathrm dx=\color{blue}\ln\left(\color{purple}\frac{3\pi^3}{32}\right)-\frac{7}{2}\cdot \color{red}\frac{\zeta(3)}{\pi^2}$$

12.$$\int_{0}^{1}\frac{x+x^2+\cdots+x^{2n+1}-(2n+1)x}{(1+x)^3\ln x}\mathrm dx=F(n)$$, $$n\ge 1$$

13.$$\int_{0}^{\infty}\frac{\sin(x)\sin(2x)\sin(4x)\cdots \sin(2^k x)}{x^{k+1}}\mathrm dx=2^{\color{purple}0.5(k^2-k+2)}\color{red}\pi$$

14.$$\int_{0}^{\infty}\frac{\sin(x)\sin(2x)}{x}\mathrm dx=\color{blue}\frac{1}{2}\ln 3$$

15.$$\int_{0}^{\infty}\frac{\sin(x)\sin(2x)\sin(3x)}{x^2}\mathrm dx=\color{red}\frac{1}{2}\ln\left(\frac{27}{4}\right)$$

16.$$\int_{0}^{\infty}\frac{x}{x^8+2x^4+1}\mathrm dx=\frac{\pi}{8}$$

17.$$\int_{0}^{\infty}\frac{x\ln x}{x^8+2x^4+1}\mathrm dx=-\frac{\pi}{16}$$

18.$$\int_{0}^{\infty}\frac{x^5\ln x}{x^8+2x^4+1}\mathrm dx=\frac{\pi}{16}$$

19.$$\int_{0}^{\infty}\frac{\sqrt[2n]{x}\ln x}{x^8+2x^4+1}\mathrm dx=-\frac{\pi}{16}\cdot \color{purple}\frac{1}{8n}\left[8n+(6n-1)\color{blue}\pi\tan\left(\frac{2n-1}{n}\cdot \frac{\pi}{8}\right)\right]\color{red}\sec\left(\frac{2n-1}{n}\cdot \frac{\pi}{8}\right)$$

20.$$\int_{0}^{\infty}\frac{\sqrt[2n]{x^{\alpha}}\ln x}{x^8+2x^4+1}\mathrm dx=-\frac{\pi}{16}\cdot \color{purple}\frac{1}{8n}\left[8n+(6n-\alpha)\color{blue}\pi\tan\left(\frac{2n-\alpha}{n}\cdot \frac{\pi}{8}\right)\right]\color{red}\sec\left(\frac{2n-\alpha}{n}\cdot \frac{\pi}{8}\right)$$

21.$$\int_{0}^{\infty}\frac{\cos\left(\frac{2x}{\pi}\right)-\cos^2\left(\frac{2x}{\pi}\right)}{x^2}\ln x\mathrm dx=\color{red}-\ln 2$$

22.$$\int_{0}^{1}\int_{0}^{1}\left(\frac{x}{1-xy}\cdot \frac{\ln x-\ln y}{\ln x+\ln y}\right)\mathrm dx \mathrm dy=1-2\color{blue}\gamma$$

23.$$\int_{0}^{\infty}\frac{4x^2}{(x^4+x^2+2)^2}\mathrm dx=\color{blue}\frac{\pi}{4}\color{green}\sqrt{5\sqrt{2}-7}$$

24.$$\int_{0}^{\infty}\sin^4(x)\ln (x)\cdot\frac{\mathrm dx}{x^2}=\color{red}\frac{\pi}{4}\color{blue}(1-\gamma)$$

25.$$\lim_{n \to \infty}\int_{-\infty}^{+\infty}\frac{\sin\left[\left(n+\frac{1}{2}\right)x\right]}{\sin\left(\frac{x}{2}\right)}\cdot\frac{\mathrm dx}{1+x^2}=\pi\cdot \color{red}\frac{e+1}{e-1}$$

26.$$\int_{0}^{1}\frac{2x^2-2x+\ln[(1-x)(1+x)^3]}{x^3\sqrt{1-x^2}}\mathrm dx=\color{red}-1$$

27.$$\int_{0}^{2\pi}\frac{x}{\phi-\cos^2(x)}\mathrm dx=2\color{red}\pi^2$$

28.$$\int_{0}^{\infty}x\sin x\ln(1-e^{-x})\mathrm dx=1-\color{brown}\frac{\pi}{2\tanh(\pi)}-\color{purple}\frac{\pi^2}{2\sinh^2(\pi)}$$

29.$$\lim_{n \to \infty}\ln n-\int_{0}^{n}\frac{e^x-x-1}{x(e^x+1)}\mathrm dx=\ln \pi-\color{purple}\gamma$$

30.$$\int_{0}^{\infty}\frac{(2x)^4}{(1-x^2+x^4)^4}\mathrm dx=\int_{0}^{\infty}\frac{(2x)^4}{(1-x^2+x^4)^3}\mathrm dx=\color{green}3\pi$$

31.$$\int_{0}^{\infty}\frac{\mathrm dx}{(1+\phi^{-2}x^2)(1+\phi^{-4}x^2)}=\color{red}\frac{\pi}{2}$$

32.$$\int_{0}^{2\pi}\frac{x\sin^3(x)}{1+\cos^2(x)}\mathrm dx=2\pi-\color{grey}\pi^2$$

33.$$\int_{0}^{\pi/2}\frac{\tan x}{\sqrt{[1+\tan^2(x)][1+\tan^6(x)]}}\mathrm dx=\int_{0}^{\pi/2}\frac{\tan^3(x)}{\sqrt{[1+\tan^2(x)][1+\tan^6(x)]}}\mathrm dx=\frac{\ln(7+4\sqrt{3})}{4\sqrt{3}}$$

34.$$\int_{-\infty}^{+\infty}\left(\frac{x}{2+2^x+2^{-x}}\right)^2\mathrm dx=\color{red}\frac{\zeta(2)-1}{3}\cdot \color{brown}\frac{1}{\ln^3(2)}$$

35.$$\int_{0}^{\pi/2}\sin^{2n+1}(x)\ln\left[\sin(x)\sin^2\left(\frac{x}{2}\right)\right]\mathrm dx=\color{blue}-4^n\sum_{j=0}^{n}\frac{(-1)^j}{(n+j+1)^2}{n \choose j}\color{green}(1+2^{-n-j})$$

36.$$\int_{0}^{\pi/2}\frac{1}{\sin^2(x)}\ln\left[\frac{\alpha+\beta\sin^2(x)}{\alpha-\beta\sin^2(x)}\right]\mathrm dx=\color{purple}\pi\cdot \frac{\color{blue}\sqrt{\alpha+\beta}-\color{red}\sqrt{\alpha-\beta}}{\sqrt{\alpha}}$$

37.$$\int_{0}^{\infty}\frac{e^x+e^{-x}-3}{(e^x+e^{-x})^2-1}\ln x \mathrm dx=\frac{\pi}{3}\cdot \color{red}\frac{\ln 2}{\sqrt{3}}$$

38.$$\int_{0}^{\infty}\sin\left(\frac{1}{4x^2}\right)\frac{\ln x}{x^2}\mathrm dx=-\color{blue}\sqrt{\frac{\pi}{2}}\cdot \color{green}\frac{\pi-2\gamma}{4}$$

39.$$\int_{0}^{1}\frac{(1-x)(x-3)}{1+x^2}\frac{\mathrm dx}{\ln x}=\ln\frac{8}{\color{green}\Gamma^4\left(\frac{3}{4}\right)}$$

40.$$\int_{-\infty}^{+\infty}\frac{(-1)^{n+1}x^{2n}+2n+1}{(1+x^2)^2}\cdot e^{-x^2}\mathrm dx=F(n)$$

$$F(n)=\color{purple}(2n+1)\sqrt{\pi}+\color{red}\frac{(-1)^n}{2^n}\sqrt{\pi}(2n-3)!!-\color{blue}\frac{\sqrt{\pi}}{2}\sum_{k=1}^{n}{n\choose k}[2(n-k)+1]\color{green}\sum_{m=0}^{k-1}{k-1 \choose m}\color{brown}\left[\frac{(-1)^{k-1}}{2^m}(2m-1)!!\right]$$

41.$$\int_{0}^{1}\frac{1-x}{1+x}\cdot\frac{2k+3+x^2}{1+x^2}\frac{\mathrm dx}{\ln x}=-\color{red}\ln(2^k \pi)$$

42.$$\int_{-\infty}^{+\infty}\frac{1}{x^2}\left(2-\frac{\sin x}{\sin\left(\frac{x}{2}\right)}\right)^{n+1}\mathrm dx={2n \choose n}\color{red}\pi$$

43.$$\int_{0}^{\infty}\frac{\ln x\ln\left(\frac{x}{1+x}\right)}{(1+x)^2}\mathrm dx=\color{blue}\zeta(2)$$

44.$$\int_{0}^{\infty}\frac{\ln x\ln\left(\frac{x}{1+x}\right)}{1+x}\mathrm dx=\color{red}\zeta(3)$$

45.$$\int_{0}^{1}\frac{\sqrt{1-x^4}}{1+x^4}\mathrm dx=\color{red}\frac{\pi}{4}$$

46.$$\int_{0}^{1}\frac{\sqrt{1-x^4}}{1+x^4}\cdot x^2\mathrm dx=\color{red}\frac{\pi}{4}-\color{blue}\frac{\color{green}\Gamma\left(\frac{2}{4}\right)\cdot \color{brown}\Gamma\left(\frac{3}{4}\right)}{\color{purple}\Gamma\left(\frac{1}{4}\right)}$$

47.$$\int_{0}^{1}\frac{2n-x-x^3-\cdots-x^{4n-1}}{1+x^2}\cdot \frac{\mathrm dx}{\ln x}=2n\ln\frac{\color{red}\Gamma\left(\frac{3}{4}\right)}{\color{purple}\Gamma\left(\frac{5}{4}\right)}-\color{blue}\ln[8^n(2n-1)!!]$$

48.$$\int_{0}^{\infty}\frac{1-e^{-\phi x}}{1+e^{+\phi x}}\cdot \frac{\mathrm dx}{x}=\color{orange}\ln\frac{\pi}{2}$$

49.$$\int_{0}^{\infty}\frac{\cosh x-1}{e^{\alpha x}-1}\cdot \frac{\mathrm dx}{x}=\frac{1}{2}\ln\frac{\pi}{\alpha\sin\left(\frac{\pi}{\alpha}\right)}$$

50.$$\int_{0}^{\infty}\frac{\gamma+\ln x}{e^x}\cdot \frac{1-\cos x}{x} \mathrm dx=\frac{1}{2}\cdot \color{blue}\frac{\pi-\ln 4}{4}\cdot \color{red}\frac{\pi+\ln 4}{4}$$

51.$$\int_{0}^{\infty}\frac{\phi^3e^{4x}-\phi^4e^{3x}-\phi^3e^{2x}+e^x+2}{(\phi e^x)^5-1}\cdot x \mathrm dx=\frac{1}{2}\color{red}\left(\frac{\pi}{5}\right)^2$$

52.$$\int_{0}^{\pi/4}\frac{\ln cot x}{\cos^{2n}(x)}\mathrm dx=\color{blue}\sum_{k=0}^{n-1}\frac{(2k+1)^2}$$

53.$$\int_{0}^{\pi/4}\frac{\sqrt{\sin(2x)}}{\cos^2(x)}\mathrm dx=2-\color{purple}\sqrt{\frac{2}{\pi}}\color{red}\Gamma^2\left(\frac{3}{4}\right)$$

54.$$\int_{0}^{1}\frac{\ln^{k}(x)}{x}\cdot \ln(1-\sqrt[n]{x})\mathrm dx=(-n)^{k+1}\color{purple}\Gamma(k+1)\zeta(n+2)$$

55.$$\int_{0}^{\infty}\cos x\ln\left(\frac{1+e^{-x}}{1-e^{-x}}\right)\mathrm dx=\sum_{n=0}^{\infty}\frac{1}{n^2+(n+1)^2}=\color{brown}\frac{\pi}{2}\tanh\left(\frac{\pi}{2}\right)$$

56. $$\int_{0}^{\infty}e^{-2x}\ln\left(\frac{1+e^{-x}}{1-e^{-x}}\right)\mathrm dx=1$$

57.$$\int_{0}^{\infty}e^{-x}\ln(x)\sin\left(\frac{1}{x}\right)\frac{\mathrm dx}{x}=-\color{blue}\frac{\pi}{2}Ker(2)$$

58.$$\int_{0}^{\pi/2}e^{-\frac{\pi}{2}\tan x}\mathrm dx=\color{purple}C_i\left(\frac{\pi}{2}\right)$$

59.$$\int_{0}^{1}\frac{1}{1+\phi x^4}\cdot \frac{\mathrm dx}{\sqrt{1-x^2}}=\color{grey}\frac{\pi}{2\sqrt{2}}$$

60.$$\int_{-1}^{+1}\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\cdot \frac{2x(x-1)+\ln[(1-x)(1+x)^3]}{x}\mathrm dx=\pi^2-\color{red}4\pi$$

61.$$\int_{0}^{1}\frac{x-x^3+x^5-x^7}{(1+x^4)\ln x}\mathrm dx=-\color{green}\ln 2$$

62.$$\int_{0}^{1}\frac{1+x+x^2+\cdots+x^{4n-1}}{(1+x^{2n})(-\ln x)^{1/\alpha}}\mathrm dx=\Gamma\left(\frac{\alpha-1}{\alpha}\right)\cdot \color{red}\sum_{k=1}^{2n}\frac{1}{k^{\frac{\alpha-1}{\alpha}}}$$

63.$$\int_{0}^{\infty}\left(2\cdot \frac{1-e^{x}}{1-e^{3x}}+\frac{1+e^{x}}{1+e^{3x}}\right)\mathrm dx=\color{blue}\ln 3$$

64.$$\int_{0}^{2e}\frac{\ln(x^2)\ln(4e-x)}{\sqrt{x(4e-x)}}\mathrm dx=\pi\color{grey}(1-\zeta(2))$$

65.$$\int_{0}^{\pi/2}\frac{\ln\cos x}{\tan x}\cdot \ln\left(\frac{\ln^2(cos x)}{\pi^2 +\ln^2(\sin x)}\right)\mathrm dx=\color{red}\zeta(2)$$

66.$$\int_{0}^{\pi/2}\frac{\ln\cos x}{\tan^2(x)}\cdot \ln\left(\frac{\ln^2(cos x)}{\pi^2 +\ln^2(\sin x)}\right)\mathrm dx=\color{purple}3\zeta(3)$$

67.$$\int_{0}^{\infty}x^{4n-1}e^{-x^n}\sin(x^n)\ln x\mathrm dx=-\color{green}\frac{3\pi}{8n^2}$$

68.$$\int_{0}^{\infty}x^{4n-1}e^{-x}\sin(x)\ln x\mathrm dx=\color{green}\pi(-1)^n\color{brown}\frac{(4n-1)!}{4^{n+1}}$$

69.$$\int_{0}^{\pi/2}\frac{\ln^2(\sin x)\ln cosx}{\sin x \cos x}\mathrm dx=-\color{blue}(\frac{\pi^2}{4!})^2$$

70.$$\int_{-\infty}^{+\infty}\sin(\cosh x)\cos(\sinh x)\mathrm dx=\color{green}\frac{\pi}{2}$$

71.$$\int_{0}^{\pi/2}\ln^2(\tan^2(x))\mathrm dx=\frac{\pi^3}{4}$$

72.$$\int_{0}^{1}\ln\left(\frac{1+x}{1-x}\right)\frac{\mathrm dx}{1+x^2}=C$$

73.$$\int_{0}^{1}\ln^{k}(x)\ln(1+x+x^2+\cdots+x^n)\frac{\mathrm dx}{x}=(-1)^k\Gamma(k+1)\color{red}\left[1-\frac{1}{(n+1)^{k+1}}\right]\color{blue}\left[k+1-\sum_{j=2}^{k+1}\zeta(j)\right]$$

74.$$\int_{0}^{\infty}\sum_{k=0}^{n}\frac{{n \choose k}e^{-(2k+1)x}}{\cosh^n(x)}\mathrm dx=\color{red}\frac{2^n}{n+1}$$

75.$$\int_{0}^{\pi/4}[4\cos^2(x)-1]\sqrt{\tan x}=\color{red}1$$

76.$$\int_{0}^{\infty}\frac{e^x-1}{e^{\alpha x}-1}\mathrm dx=-\frac{1}{\alpha}\color{red}\left(\gamma+\psi\left(1-\frac{1}{\alpha}\right)\right)$$

77.$$\int_{0}^{\infty}\frac{x}{e^x-1}\ln\left(\frac{e^x+1}{e^x-1}\right)\mathrm dx=\left(\frac{\pi}{2}\right)\color{blue}\ln 2$$

78.$$\int_{0}^{1}\left(\frac{x^u}{\ln x}+\frac{x^v}{1-x}\right)\mathrm dx=\color{blue}\gamma-H_v+\color{red}\ln(u+1)$$

79.$$\int_{0}^{\pi/4}\sin(2x)\ln(\ln^2(\cot x))\mathrm dx=ln\frac{\pi}{4}-\color{orange}\gamma$$

80.$$\int_{0}^{\infty}\ln x\sin\left[\pi\left(x^2-\frac{1}{4}\right)\right]\mathrm dx=\color{red}\frac{\pi}{8}$$

81.$$\int_{0}^{\infty}\sum_{i=0}^{j}(-1)^{i+j}{j \choose i}\frac{e^{-x^{i+\alpha}}}{x}\mathrm dx=-\color{green}\gamma(-1)^j\frac{j!}{\color{blue}\alpha(\alpha+1)(\alpha+2)\cdots (\alpha+j)}$$

82.$$1-2\int_{0}^{\infty}\frac{\arctan x}{e^{\pi x}-1}\mathrm dx=\ln 2$$

83.$$\int_{0}^{\pi}\frac{\sin^{2k-1}(x)}{1+\cos^2(x)}\mathrm dx=2^{k-2}\pi-\color{purple}\frac{1}{2}\sum_{j=1}^{k-1}\frac{2^{j+k}}{j{2j \choose j}}$$

84.$$\int_{0}^{1}x^n\left(\frac{1}{\ln x}+\frac{1}{1-x}\right)\mathrm dx=\gamma+\ln(1+n)-H_n$$

85.$$\int_{0}^{\infty}\frac{\sin^2(\beta x)}{x^2}\frac{\mathrm dx}{e^{2\alpha x}}=\color{green}\beta\arctan\left(\frac{\beta}{\alpha}\right)-\color{red}\frac{\alpha}{2}\ln\left(\frac{\alpha^2+\beta^2}{\alpha^2}\right)$$

86.$$\int_{0}^{1}\frac{1-x^{\sqrt{5}}}{(1-x^{\phi})^{\phi}}\mathrm dx=\color{orange}\phi^2$$

87.$$\int_{0}^{\infty}\frac{\ln x}{x}\left(\frac{1+x+e^x}{e^x+e^{2x}}-\frac{1}{\sqrt{1+x^2}}\right)\mathrm dx=\color{blue}\frac{\ln^2(2)-\gamma^2}{2}$$

88.$$\int_{0}^{\pi}\frac{\sin^5(x)}{1+\cos^3(x)}\mathrm dx=\color{blue}\ln 3$$

89.$$\int_{-\infty}^{+\infty}\frac{1}{x^2}\left(1-\frac{\sin x}{x}\right)^n\mathrm dx=\color{red}\frac{\pi}{n+1}$$

90.$$\int_{0}^{1}\frac{\ln^{2k+1}(x)\ln(1-x)}{x}\mathrm dx=\color{red}\Gamma(2k+2)\zeta(2k+3)$$

91.$$\int_{-1}^{+1}\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{1-x+2x^3}{1+x-2x^3}\right)\mathrm dx=2\pi\color{green}\arccot(2\sqrt{\phi^3})$$

92.$$\int_{0}^{1}\frac{1}{x^2}e^{-x^2}\sin(x^2)\mathrm dx=\sqrt{\pi}\color{purple}(\sqrt{\phi}-1)$$

93.$$\int_{0}^{1}\frac{2x+\ln\left(\frac{1-x}{1+x}\right)}{x^3\sqrt{1-x^2}}\mathrm dx=-\color{red}\left(\frac{\pi}{2}\right)^2$$

94.$$\int_{0}^{1}\frac{2x+\ln\left(\frac{1-x}{1+x}\right)}{x^2\sqrt{1-x^2}}\mathrm dx=-\color{grey}2$$

95.$$\int_{0}^{\infty}\left(\frac{1}{1+nx^n}-\frac{1}{e^{nx^n}}\right)\frac{\mathrm dx}{x^{n+1}}=1-\color{brown}\gamma$$

96.$$\int_{-\infty}^{+\infty}\frac{b}{1+2ax+(cx)^2}\cdot \frac{\mathrm dx}{1-2ax+(cx)^2}=\color{red}\frac{\pi}{2}$$

97.$$\int_{0}^{1}\sqrt{\frac{x}{1-x}}\ln\left(\frac{x}{1-x}\right)\mathrm dx=\pi$$

98.$$\int_{0}^{\infty}e^{-\sqrt{x}}\ln\left(1+\frac{1}{\sqrt{x}}\right)\mathrm dx=\color{blue}2\gamma$$

99.$$\int_{-\infty}^{+\infty}xe^{-\pi x^2\left(\frac{\alpha+x}{\beta+x}\right)^2}\mathrm dx=\beta-\alpha$$

100.$$\int_{0}^{1}\int_{0}^{1}\frac{1-x}{1-xy}\frac{x}{\ln(xy)}\mathrm dx \mathrm dy=\frac{1}{2}\color{red}\ln\frac{1}{2}$$

101.$$\int_{0}^{1}\frac{x^{2n}-x}{1+x}\frac{\mathrm dx}{ln x}=\color{purple}\ln\left(\frac{\pi}{2}\cdot\frac{(2n)!!}{(2n-1)!!}\right)$$

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1.$$\int_{0}^{1}\frac{x^{\alpha}\ln^2(x)}{x\sqrt{1-x^{\alpha}}}\mathrm dx=\color{red}\left(\frac{2}{\alpha}\right)^3\left[\color{blue}2(1-\ln 2)-\color{green}\eta(2)+\color{purple}\ln^2(2)\right]$$

2.$$\int_{0}^{\infty}\frac{\ln x}{(1+\alpha_0 x)(1+\alpha_1 x)}\mathrm dx=-\frac{\ln^2(\alpha_0)-\ln^2(\alpha_1)}{4}$$

3.$$\int_{0}^{\infty}\frac{\ln x}{(1+\alpha_0 x)(1+\alpha_1 x)(1+\alpha_2 x)}\mathrm dx=-\frac{\alpha_0\ln^2(\alpha_0)-2\alpha_1\ln^2(\alpha_1)+\alpha_2\ln^2(\alpha_2)}{16}$$

4.$$\int_{0}^{\infty}\frac{\ln x}{(1+\alpha_0 x)(1+\alpha_1 x)(1+\alpha_2 x)(1+\alpha_3 x)}\mathrm dx=-\frac{\alpha_0^2\ln^2(\alpha_0)-3\alpha_1^2\ln^2(\alpha_1)+3\alpha_2^2\ln^2(\alpha_2)-\alpha_3^2\ln^2(\alpha_3)}{96}$$

5.$$\int_{0}^{\infty}\frac{\ln x}{(1+\alpha_0 x)(1+\alpha_1 x)(1+\alpha_2 x)(1+\alpha_3 x)(1+\alpha_4 x)}\mathrm dx=-\frac{\alpha_0^3\ln^2(\alpha_0)-4\alpha_1^3\ln^2(\alpha_1)+6\alpha_2^3\ln^2(\alpha_2)-4\alpha_3^3\ln^2(\alpha_3)+\alpha_4^3\ln^2(\alpha_4)}{768}$$

5.$$\int_{0}^{\infty}\frac{\ln x}{(1+\alpha_0 x)(1+\alpha_1 x)(1+\alpha_2 x)\cdots(1+\alpha_n x)}\mathrm dx=-\color{red}\frac{1}{2^{n+1}n!}\color{blue}\sum_{j=0}^{n}(-1)^j{n\choose j}\alpha_j^{n-1}\ln^2(\alpha_j)$$

6.$$\int_{0}^{\infty}\frac{\ln^2(x)}{(1+\alpha x)(1+\beta x)}\mathrm dx=-\color{blue}\frac{\ln^3(\alpha)-\ln^3(\beta)}{6}+\color{red}\zeta(2)\color{purple}\ln\left(\frac{\alpha}{\beta}\right)$$

7.$$\int_{0}^{\infty}\frac{\ln^2(x)}{(1+\alpha_0 x)(1+\alpha_1 x)(1+\alpha_2 x)\cdots(1+\alpha_n x)}\mathrm dx=G(2,\alpha_n)$$

8.$$\int_{0}^{\infty}\frac{\ln x}{[(1+\alpha x)(1+(1+\alpha)x)]^2}\mathrm dx=-\alpha(\alpha+1)\left[\ln^2(\alpha)-\ln^2(\alpha+1)+\ln\left[\alpha^{\alpha}(\alpha+1)^{\alpha+1}\right]\right]$$

9.$$\int_{0}^{\pi/2}\frac{\tan x}{(\alpha+\beta\tan^2 x)^2}\mathrm dx=\color{blue}\frac{\beta-\alpha\left(1+\ln\frac{\beta}{\alpha}\right)}{\color{red}2\alpha(\beta-\alpha)^2}$$

10.$$\int_{0}^{\pi/2}\frac{\tan x}{(\alpha+\beta\tan^2 x)^3}\mathrm dx=\color{blue}\frac{\beta^2-4\alpha\beta+\alpha^2\left(3+2\ln\frac{\beta}{\alpha}\right)}{\color{red}4\alpha^2(\beta-\alpha)^3}$$

11.$$\int_{0}^{\pi/2}\frac{\tan x}{(\alpha+\beta\tan^2 x)^4}\mathrm dx=\color{blue}\frac{2\beta^3-9\alpha\beta^2+18\alpha^2\beta -\alpha^3\left(11+6\ln\frac{\beta}{\alpha}\right)}{\color{red}12\alpha^3(\beta-\alpha)^4}$$

12.$$\int_{0}^{\pi}\frac{\sin x}{\alpha^2+\beta^2-2\alpha\beta\cos x}\mathrm dx=\color{blue}\frac{1}{\alpha\beta}\ln\left(\color{green}\frac{\beta+\alpha}{\beta-\alpha}\right)$$

13.$$\int_{0}^{\pi/2}\frac{\sin x}{\alpha^2+\beta^2-2\alpha\beta\cos x}\mathrm dx=\color{green}\frac{1}{2\alpha\beta}\ln\left(\frac{\beta^2+\alpha^2}{(\beta-\alpha)^2}\right)$$

14.$$\int_{0}^{2\pi}\frac{\sin^2(x)}{\alpha^2+\beta^2-2\alpha\beta\cos x}\mathrm dx=\color{red}\frac{\pi}{\beta^2}$$

15.$$\int_{0}^{\pi}\frac{\sin x}{(\alpha^2+\beta^2-2\alpha\beta\cos x)^2}\mathrm dx=\color{purple}\frac{2}{(\beta-\alpha)^2}$$

16.$$\int_{-\infty}^{+\infty}\frac{x^2}{x^4+\alpha x+\beta^2}\mathrm dx=\color{green}\frac{\pi}{\sqrt{2\beta+\alpha}}$$

17.$$\int_{0}^{1}\frac{\ln(1+x)\ln^2(x)}{(1+x)^2}\mathrm dx=\color{red}\zeta(2)-\color{blue}\frac{5}{4}\zeta(3)$$

18.$$\int_{0}^{1}\frac{\ln(1-x)\ln^2(x)}{(1-x)^2}\mathrm dx=2\color{red}\zeta(2)-\color{purple}{4}\zeta(3)$$

19.$$\int_{0}^{\infty}\arctan\left(\frac{\alpha}{\beta+x}\right)\frac{\mathrm dx}{(\beta+x)^2}=\frac{1}{\beta}\color{red}\arctan\left(\frac{\alpha}{\beta}\right)-\color{purple}\frac{1}{2\alpha}\ln\left(\frac{\alpha^2+\beta^2}{\beta^2}\right)$$

20.$$\int_{0}^{\infty}\arctan\left(\frac{\alpha}{\beta+x}\right)\frac{\mathrm dx}{(\beta+x)^3}=\color{red}\frac{1}{2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\right)\color{purple}\arctan\left(\frac{\alpha}{\beta}\right)-\color{blue}\frac{1}{2\alpha\beta}$$

21.$$\int_{0}^{\infty}\arctan\left(\frac{\alpha}{\beta+x}\right)\frac{\mathrm dx}{(\beta+x)^4}=\frac{1}{6\beta^3}\color{red}\arctan\left(\frac{\alpha}{\beta}\right)+\color{purple}\frac{1}{6\alpha^3}\ln\left(\frac{\alpha^2+\beta^2}{\beta^2}\right)-\color{brown}\frac{1}{6\alpha\beta^2}$$

22.$$\int_{0}^{\infty}\arctan\left(\frac{\alpha}{\beta+x}\right)\frac{\mathrm dx}{(\beta+x)^5}=\color{red}\frac{1}{4}\left(\frac{1}{\beta^4}-\frac{1}{\alpha^4}\right)\color{purple}\arctan\left(\frac{\alpha}{\beta}\right)+\color{blue}\frac{1}{4\alpha^3\beta}-\color{green}\frac{1}{12\alpha\beta^3}$$

23.$$\int_{-\infty}^{+\infty}\frac{x \arctan x}{(\alpha^2+x^2)^2}\mathrm dx=\color{red}\frac{\pi}{2}\cdot \color{blue }\frac{1}{\alpha+\alpha^2}$$

23.$$\int_{-\infty}^{+\infty}\frac{x \arctan\left(\frac{\alpha}{\gamma+x}\right)}{(\beta^2+x^2)^2}\mathrm dx=\color{red}\frac{\pi}{2}\cdot\color{blue}\frac{\alpha}{\beta}\cdot \color{green}\frac{\beta(\alpha+\beta)-\gamma^2}{\color{purple}(\gamma^2+\alpha^2)[\gamma^2+(\alpha+\beta)^2]}$$

24.$$\int_{0}^{1}\left(\frac{x^n+\alpha}{\ln x}-\frac{(\alpha+1)\ln x}{x^2-1}\right)\mathrm dx=\color{red}(\alpha+1)\color{blue}(\gamma+1)+\color{purple}\ln(n+1)$$

25.$$\int_{0}^{1}\left(\frac{x^n+\alpha}{\ln x}-\frac{\alpha+1}{x-1}\right)\cdot \frac{\ln x}{x-1}\mathrm dx=\color{red}(\alpha+1)+\color{blue}H_n$$

26.$$\int_{0}^{1}\left(\frac{x^n+\alpha}{\ln x}-\frac{(\alpha+1)\ln x}{(1-x)^2}\right)\cdot\left(\frac{\ln x}{x-1}\right)^m\mathrm dx=H_n+\color{red}(\alpha+1)\color{blue}G(m)$$

$$G(1)=\zeta(2)+\frac{3}{2}$$

$$G(2)=2\zeta(3)+\zeta(2)+\frac{5}{6}$$

$$G(3)=4\zeta(4)+11\zeta(3)+5\zeta(2)+\frac{5}{12}$$

27.$$\int_{0}^{1}x^n\left(\frac{1}{\ln x}-\frac{\ln x}{(1-x)^2}\right)\mathrm dx=\color{red}1+\gamma-\color{blue}n\zeta(2)+\color{brown}\ln(n+1)+\color{purple}\sum_{j=0}^{n}H_j^2$$

28.$$\int_{0}^{1}\sqrt{x}\left(\frac{1}{\ln x}-\frac{\ln x}{(1-x)^2}\right)\mathrm dx=\color{red}1+\gamma-\left(\frac{\pi}{2}\right)^2+\color{green}\ln 3!$$

29.$$\int_{0}^{1}x^n\left(\frac{1}{\ln x}-\frac{\ln x}{(1-x)^2}\right)^2\mathrm dx=\color{red}(1+n)\gamma+(n+1)\color{brown}\ln(n+1)+G(n)$$

30.$$\int_{0}^{1}x^n\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)\left(\frac{\ln x}{x-1}\right)^{n+1}\mathrm dx=\color{blue}\frac{1}{1+n}$$

31.$$\int_{0}^{1}x^n\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)\left(\frac{\ln x}{x-1}\right)^{n+1}\ln x\mathrm dx=\color{red}\sum_{j=1}^{n}(-1)^ja_j\zeta(j+1)$$

32.$$\int_{0}^{\pi/2}\sin x\arctan(\alpha \tan x)\mathrm dx=\frac{\alpha}{\sqrt{\alpha^2-1}}\arccos\left(\frac{1}{\alpha}\right)$$

33.$$\int_{0}^{\pi/2}\sin(2x)\arctan(\alpha \tan x)\mathrm dx=\frac{\pi}{2}\cdot \frac{\alpha}{1+\alpha}$$

34.$$\int_{0}^{\pi/2}\sin(4x)\arctan(\alpha \tan x)\mathrm dx=-\frac{\pi}{2}\cdot \frac{\alpha}{(1+\alpha)^2}$$

35.$$\int_{0}^{\pi/2}\cos x\arctan(\alpha \tan x)\mathrm dx=\frac{\pi}{2}-\frac{1}{\sqrt{\alpha^2-1}}\operatorname{arccosh}\left(\frac{1}{\alpha}\right)$$

36.$$\int_{0}^{\pi/2}\cos(2x)\arctan(\alpha \tan x)\mathrm dx=\frac{\alpha}{1-\alpha^2}\cdot\ln\alpha$$

37.$$\int_{0}^{1}x^{n-1}\arctan\left(\frac{1-x^n}{1+x^n}\right)\mathrm dx=-\color{grey}\frac{1}{48n^2}(\pi^2+24\ln 2)$$

38.$$\int_{0}^{1}\frac{1-x^n}{1+x^n}\cdot \frac{x^{n-1}}{\ln x}\mathrm dx=\color{brown}\ln\left(\frac{2}{\pi}\right)$$

39.$$\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^2}{\ln x}\mathrm dx=\color{orange}\ln\left(\frac{8}{3\pi}\right)$$

40.$$\int_{0}^{1}\frac{1-x^2}{1+x^2}\cdot \frac{x^3}{\ln x}\mathrm dx=\color{purple}\ln\left(\frac{\pi}{4}\right)$$

41.$$\int_{0}^{1}\frac{1-x^4}{1+x^4}\cdot \frac{x^5}{\ln x}\mathrm dx=\color{red}\ln\left(\frac{4}{3}\right)+2\color{blue}\ln\left[\frac{\Gamma(5/4)}{\Gamma(3/4)}\right]$$

42.$$\int_{0}^{1}\frac{x^2}{1+x}\ln(1+x^2)\mathrm dx=3\left[\frac{1}{2}+\left(\frac{\ln 2}{2}\right)\right]-\frac{\pi}{3\cdot 2^4}(\pi+24)$$

43.$$\int_{0}^{\infty}\frac{\ln x}{x^2+2x+5}\mathrm dx=\frac{1}{4}\color{red}\ln(5)\color{blue}\arctan(2)$$

44.$$\int_{0}^{\infty}\frac{\ln x}{x^{\alpha}+1}\mathrm dx=-\frac{\pi^2}{\alpha^2}\cot\left(\frac{\pi}{\alpha}\right)\csc\left(\frac{\pi}{\alpha}\right)$$

45.$$\int_{0}^{\infty}\frac{\ln x}{x^{\alpha}+\beta}\mathrm dx=-\frac{\pi}{\alpha^2}\beta^{\frac{1}{\alpha}-1}\csc\left(\frac{\pi}{\alpha}\right)\left[\pi\cot\left(\frac{\pi}{\alpha}\right)-\ln\beta\right]$$

46.$$\int_{0}^{\infty}\frac{\ln x}{\gamma x^{\alpha}+\beta}\mathrm dx=-\frac{\pi}{\alpha^2}\gamma^{-\frac{1}{\alpha}}\beta^{\frac{1}{\alpha}-1}\csc\left(\frac{\pi}{\alpha}\right)\left[\pi\cot\left(\frac{\pi}{\alpha}\right)+\ln\left(\frac{\gamma}{\beta}\right)\right]$$

47.$$\int_{0}^{1}\frac{\ln x}{(1-x^n)^{1/n}}\mathrm dx=\color{orange}\frac{\pi}{n^2}\csc\left(\frac{\pi}{n}\right)\left(\psi_0\left(\frac{1}{n}\right)+\gamma\right)$$

48.$$\int_{0}^{\pi}\sin^{\alpha}\sin(\alpha x)\mathrm dx=-\color{purple}\frac{\pi}{2^{\alpha}}$$; $$\alpha=2n+1$$; $$n\ge0$$

49.$$\int_{0}^{\pi}x\sin^{\alpha}\sin(\alpha x)\mathrm dx=\color{blue}(-1)^n\frac{\pi^2}{2^{\alpha+1}}$$,$$\alpha=2n+1$$; $$n\ge0$$

50.$$\int_{0}^{\pi}\cos^{\alpha}\left(\frac{\alpha}{2}x\right)\cos(\alpha x)\mathrm dx=\frac{(-1)^n}{4}\color{red}\frac{(\alpha+1)!!}{(\alpha+2)!!}$$, $$\alpha=2n+1$$; $$n\ge0$$

51.$$\int_{-1}^{1}\frac{\sqrt{1-x^2}\ln(1-x^2)}{x^2}\mathrm dx=\color{blue}2\pi\color{red}(\ln 2-1)$$

52.$$\int_{0}^{1}\frac{x(1-x)}{1+x+x^2}\frac{\mathrm dx}{\ln x}=\color{red}\ln\left(\frac{\sqrt{27}}{2\pi}\right)$$

53.$$\int_{0}^{1}\frac{(1-x)^3}{1+x+x^2}\frac{\mathrm dx}{\ln x}=\color{blue}\ln\left(\frac{4\pi^3}{27\sqrt{27}}\right)$$

54.$$\int_{0}^{1}\frac{x(1-x)^2}{1+x+x^2}\frac{\mathrm dx}{\ln x}=\color{red}\ln(18)-\color{green}3\ln\Gamma\left(\frac{1}{3}\right)$$

55.$$\int_{0}^{1}\frac{1-x+x^2}{1+x+x^2}\frac{1-x}{\ln x}\mathrm dx=\color{red}\ln\left[\frac{2}{3}\cdot \left(\frac{\pi}{3}\right)^2\right]$$

56.$$\int_{\pi/4}^{\pi/2}\sin[4(2n+1)x]\ln(1+\cot x)\mathrm dx=-\color{blue}\frac{\ln 2}{4(2n+1)}$$, $$n\ge0$$

57.$$\int_{\pi/4}^{\pi/2}\sin(8x)\ln(1+\cot x)\mathrm dx=\color{purple}\frac{1}{12}$$

58.$$\int_{\pi/4}^{\pi/2}\ln(1+\cot x)\mathrm dx=\color{green}\frac{\pi \ln 2}{8}$$

59.$$\int_{0}^{\pi/2}\sin(2x)\ln(1+\cot x)\mathrm dx=\color{green}\frac{\pi}{4}$$

60.$$\int_{0}^{\pi/2}\sin(4x)\ln(1+\cot x)\mathrm dx=\color{orange}\frac{1}{2}$$

61.$$\int_{0}^{\pi/2}\sin(8x)\ln(1+\cot x)\mathrm dx=\color{yellow}\frac{1}{3}$$

62.$$\int_{0}^{\pi/4}\sin(8x)\ln(1+\cot x)\mathrm dx=\color{brown}\frac{1}{4}$$

63.$$\int_{0}^{\pi/4}\sin(16x)\ln(1+\cot x)\mathrm dx=\color{grey}\frac{1}{6}$$

64.$$\int_{0}^{1}\int_{0}^{1}\frac{2+x^2-y^2}{2-x^2-y^2}\mathrm dx \mathrm dy=2C$$

65.$$\int_{0}^{\pi/2}x \ln(1+\cot x)\mathrm dx=\color{brown}\frac{2^2\pi C-7\zeta(3)+\pi^2\ln 2}{4^2}$$

66.$$\int_{0}^{\pi/4}x \ln(1+\cot x)\mathrm dx=\color{purple}\frac{2^3\pi C-7\zeta(3)+\pi^2\ln 2}{4^3}$$

67.$$\int_{\alpha}^{\beta}\frac{\ln x}{\sqrt{(\alpha-x)(x-\beta)}}\mathrm dx=\color{red}2\pi\ln\left(\color{blue}\frac{\sqrt{\alpha}+\sqrt{\beta}}{2}\right)$$

68.$$\int_{0}^{1}\frac{1-x}{1+x}\cdot \ln^{n}(x)\mathrm dx=(-1)^n\color{blue}\Gamma(n+1)[\color{red}2\eta(n+1)-1]$$

69.$$\int_{0}^{1}\frac{1-x^2}{1+x^2}\cdot \ln^{n}(x)\mathrm dx=(-1)^n\color{blue}\Gamma(n+1)[\color{purple}2\beta(n+1)-1]$$

70.$$\int_{\pi/4}^{\pi/2}\frac{\ln\tan x}{\cos(2x)}\mathrm dx=-\color{red}\lambda(2)$$

71.$$\int_{\pi/4}^{\pi/2}\frac{\ln\tan x}{\cos(x/2)}\mathrm dx=-\color{purple}\frac{\pi^2}{9\ln 2}$$

72.$$\int_{\pi/4}^{\pi/2}\frac{\ln^{n}\tan x}{\cos(x/2)}\mathrm dx=-\color{brown}\Gamma(n+1)\lambda(n+1)$$

73.$$\int_{0}^{\pi/4}\frac{\cos[(4n+1)x]}{\cos x}\mathrm dx=\frac{\pi}{4}-\sum_{j=1}^{n}\frac{1}{2j-1}$$, $$n\ge0$$

74.$$\int_{0}^{\pi/4}\frac{\cos(2nx)}{\cos^{2n+2}(x)}\mathrm dx=\color{red}\frac{2^n}{2n+1}$$

75.$$\int_{0}^{\infty}\frac{x\ln(1+x^2)}{\sinh(\pi x)}\mathrm dx=\color{red}18\ln A+\color{blue}\frac{1}{3}\ln 2-\color{green}\ln \pi -\frac{3}{2}$$

76.$$\int_{-\infty}^{+\infty}\frac{x}{1+x^4}\sin\left(\alpha\sqrt{2} x\right)\mathrm dx=\color{blue}\frac{\pi}{e}\color{red}\sin(\alpha)$$

77.$$\int_{-\infty}^{+\infty}\frac{x^2}{1+x^4}\sin^2\left(\alpha\sqrt{2} x\right)\mathrm dx=\color{green}\frac{\pi}{2\sqrt{2}}\cdot \frac{\color{blue}e^{2\alpha}+\color{purple}\sin(2\alpha)-\cos(2\alpha)}{\color{red}e^{2\alpha}}$$

78.$$\int_{0}^{\infty}\left(\frac{\tanh x}{x^2}-\frac{\operatorname{sech^2(x)}}{x}\right)\mathrm dx=\color{blue}1$$

79.$$\int_{0}^{\infty}\left(\frac{\tanh x}{x^2}-\frac{\operatorname{sech^2(x)}}{x}\right)\frac{\mathrm dx}{\sinh x}=\color{red}\ln 2$$

80.$$\int_{-\infty}^{+\infty}e^{i \alpha x}\left(\frac{1}{\beta+ix}+\frac{1}{\beta-ix}\right)^2\mathrm dx=\color{brown}2\pi e^{-\alpha\beta}\cdot\color{green} \frac{\alpha\beta+1}{\beta}$$

81.$$\int_{-\infty}^{+\infty}e^{i \alpha x}\left(\frac{1}{\beta+ix}-\frac{1}{\beta-ix}\right)^2\mathrm dx=\color{red}2\pi e^{-\alpha\beta}\cdot \color{blue}\frac{\alpha\beta-1}{\beta}$$

82.$$\int_{-\infty}^{+\infty}e^{i \alpha x}\left(\frac{1}{\beta+ix}-\frac{1}{\beta-ix}\right)\mathrm dx=\color{red}2\pi e^{-\alpha\beta}$$

83.$$\int_{-\infty}^{+\infty}e^{i \alpha x}\left(\frac{1}{(\beta+ix)^n}+\frac{1}{(\beta-ix)^n}\right)\mathrm dx=\color{red}\pi e^{-\alpha\beta}\cdot \color{purple}\frac{2^n}{\Gamma(n)}$$

84.$$\int_{-\infty}^{+\infty}\frac{e^{i \alpha x}}{\beta^2+x^2}\mathrm dx=\color{green}\frac{\pi e^{-\alpha\beta}}{\beta}$$

85.$$\int_{0}^{\infty}\frac{\mathrm dx}{(x+\alpha)(x+\beta)^2}=\frac{1}{(\beta-\alpha)^2}\color{purple}\ln\left(\frac{\beta}{\alpha}\right)-\color{red}\frac{1}{\beta(\beta-\alpha)}$$

86.$$\int_{-\infty}^{+\infty}\frac{x^2+2}{x^4+4}\cos(\alpha x)\mathrm dx=\pi \color{blue}e^{-\alpha}\color{red}\cos(\alpha)$$

87.$$\int_{0}^{\pi}\frac{\cos(n x)}{\alpha^2+2\alpha \cos x+1}\mathrm dx=\color{blue}\frac{(-\alpha)^{n}}{1-\alpha^2}\color{red}\pi$$

88.$$\int_{0}^{\pi}\frac{\sin x}{\alpha^2+2\alpha \sin x+1}\mathrm dx=\color{red}\frac{\pi-2(1+\alpha^2)\arccot(\alpha)}{\color{blue}\alpha(1-\alpha^2)}$$

89.$$\int_{0}^{\pi}\frac{\sin x}{\alpha^2+2\alpha \cos x+1}\mathrm dx=\color{red}\frac{2}{\alpha}\operatorname{arctanh}(\alpha)$$

90.$$\int_{0}^{\pi/2}\frac{\cos x}{(\alpha^2+2\alpha\sin x+1)^{n+1}}\mathrm dx=\color{green}\frac{1}{2\alpha n}\left[\color{red}\frac{1}{(1+\alpha^2)^n}-\color{blue}\frac{1}{(1+\alpha)^{2n}}\right]$$, $$n\ge1$$

91.$$\int_{0}^{\pi/n}\frac{\cos x}{\alpha^4+2\alpha^2\cos(2x)+1}\mathrm dx=\frac{1}{2\alpha(1+\alpha^2)}\color{purple}\operatorname{arctanh}\left(\color{blue}\frac{2\alpha}{1+\alpha^2}\cdot\color{red}\sin\left(\frac{\pi}{n}\right)\right)$$

92.$$\int_{0}^{\pi/n}\frac{\cos^2(x)}{\alpha^4+2\alpha^2\cos(2x)+1}\mathrm dx=\frac{\pi\left[(n+2)\alpha^2-(n-2)\right]+2n(\alpha^2-1)\arctan\left(\color{blue}\frac{\alpha^2+1}{\alpha^2-1}\cdot \cot\color{red}\left(\frac{\pi}{n}\right)\right)}{\color{purple}8n\alpha^2(1+\alpha^2)}$$

93.$$\int_{0}^{\pi}\frac{\cos^2(x)}{\alpha^4+2\alpha^2\cos(2x)+1}\mathrm dx=\frac{\pi}{2}\cdot \frac{1}{1+\alpha^2}$$

94.$$\int_{0}^{\pi/2}\frac{\cos^2(x)}{\alpha^4+2\alpha^2\cos(2x)+1}\mathrm dx=\frac{\pi}{4}\cdot\frac{1}{1+\alpha^2}$$

95.$$\int_{0}^{1}\frac{\ln^{n}(1-x^{\alpha})}{x}\mathrm dx=\frac{1}{\alpha}\color{blue}\Gamma(n+1)\color{red}\zeta(n+1)$$

96.$$2\int_{-\infty}^{+\infty}\frac{ke^x+1}{\pi^2+(e^x-x+1)^2}\cdot \frac{(e^x+1)^2}{\pi^2+(e^x+x+1)^2}x \mathrm dx=\color{red}k$$

97.$$\int_{0}^{\infty}\frac{\mathrm dx}{1+x^2+x^4}=\int_{0}^{\infty}\frac{\mathrm dx}{1+x^4+x^8}$$

98.$$\int_{0}^{1}\frac{(1-x)(1-2x^{\phi})+\phi(x-x^{\phi})}{(1-x)^2}\cdot \left(\frac{1-x^{\phi}}{1-x}\right)^{1/\phi}\mathrm dx=\color{green}\phi^{\phi}$$

99.$$\int_{0}^{1}\cos(\ln x)\frac{\mathrm dx}{1+x^2}=\color{red}\frac{\pi}{4}\cdot \color{blue}\frac{1}{\cosh\left(\frac{\pi}{2}\right)}$$

100.$$\int_{0}^{1}\frac{\cos(\ln x)\ln^2(x)}{1+x^2}\mathrm dx=\beta(3)\color{red}\frac{3-\cosh(\pi)}{\cosh^3\left(\color{blue}\frac{\pi}{2}\right)}$$

---

1.$$\int_{\alpha^2}^{\beta^2}\arccos\left(\frac{x}{\sqrt{(\alpha^2+\beta^2)x-(\alpha\beta)^2}}\right)\mathrm dx=\color{brown}\frac{(\alpha+\beta)^2}{\alpha^2+\beta^2}\cdot\color{green}\frac{\pi}{4}$$

2.$$\int_{0}^{\infty}\frac{\ln(x)\arccot(x)}{\sqrt[\alpha]{x}}\mathrm dx=\color{blue}\frac{\pi}{4}\cdot \color{red}\frac{\alpha}{(\alpha-1)^2}\cdot \color{purple}\frac{(\alpha-1)\pi\cot\left(\frac{\pi}{2\alpha}\right)-2\alpha}{\color{brown}\sin\left(\frac{\pi}{2\alpha}\right)}$$

3.$$\int_{0}^{\infty}\frac{\ln^2(x)\arccot(x)}{\sqrt[\alpha]{x}}\mathrm dx=G(\alpha)$$

$$G(\alpha)=\frac{\alpha \pi}{(\alpha-1)^3\sin^3\left(\frac{\pi}{2\alpha}\right)}\cdot \frac{3\pi^2+\alpha\pi^2[8\alpha+3(\alpha-2)]-4(\alpha-1)\alpha \pi\sin(\pi/\alpha)+[\pi^2(\alpha-1)^2-8\alpha^2]\cos(\pi/\alpha)}{16}$$

4.$$\int_{0}^{\infty}\frac{\ln(1+x^2)\arccot(x)}{x}\mathrm dx=\color{red}\frac{\pi^3}{12}$$

5.$$\int_{0}^{\infty}\frac{\ln(1+x^2)\arccot(x)}{x^2}\mathrm dx=\color{purple}\zeta(2)$$

6.$$\int_{0}^{\infty}\frac{\ln(1+x^{2\alpha})\arccot(x)}{x}\mathrm dx=\color{red}G(\alpha)\color{blue}\pi^3$$

7.$$\int_{0}^{\infty}\frac{\ln^{2\alpha}(x)\ln(1+x)\arccot(x)}{x}\mathrm dx=\color{red}G(\alpha)\pi^{2\alpha+3}$$

8.$$\int_{0}^{1}\frac{\ln(1+x^2)}{x\sqrt{x(1-x)}}\mathrm dx=\color{blue}2\pi\left(\sqrt{\color{red}2(1+\color{green}\sqrt{2})}-2\right)$$

9.$$\int_{0}^{\pi/\alpha}\cos\left(\frac{x}{2}\right)\ln\left[\tan\left(\frac{x}{2}\right)\right]\mathrm dx=2\sin\left(\frac{\pi}{2\alpha}\right)\ln\left[\tan\left(\frac{\pi}{2\alpha}\right)\right]-4\operatorname{arctanh}\left[\tan\left(\frac{\pi}{4\alpha}\right)\right]$$

10.$$\int_{0}^{\pi/3}\cos\left(\frac{x}{2}\right)\ln\left[\tan\left(\frac{x}{2}\right)\right]\mathrm dx=-\frac{3}{2}\ln 3$$

11.$$\int_{0}^{\pi/2}\cos^2\left(\frac{x}{2}\right)\ln\left[\cot\left(\frac{x}{2}\right)\right]\mathrm dx=\beta(1)+\beta(2)$$

12.$$\int_{0}^{\pi}\cos^2\left(\frac{x}{2}\right)\ln\left[\cot\left(\frac{x}{2}\right)\right]\mathrm dx=\color{red}\frac{\pi}{2}$$

13.$$\int_{0}^{\pi}\cos^4\left(\frac{x}{2}\right)\ln\left[\cot\left(\frac{x}{2}\right)\right]\mathrm dx=\color{red}\frac{\pi}{2}$$

14.$$\int_{0}^{\pi/3}\cos^3\left(\frac{x}{2}\right)\ln\left[\cot\left(\frac{x}{2}\right)\right]\mathrm dx=\color{red}\frac{1}{3}+\frac{9}{8}\ln 3$$

15.$$\int_{0}^{\pi/\alpha}\frac{1}{2}\cdot\sin\left(\frac{x}{2}\right)\ln\left[\cot\left(\frac{x}{2}\right)\right]\mathrm dx=\ln 2+\ln\left[\tan\left(\frac{\pi}{4\alpha}\right)\right]+\cos\left(\frac{\pi}{2\alpha}\right)\ln\left[\cot\left(\frac{\pi}{2\alpha}\right)\right]$$

16.$$\int_{0}^{\pi/2}\sin^2\left(\frac{x}{2}\right)\ln\left[\tan\left(\frac{x}{2}\right)\right]\mathrm dx=\beta(1)-\beta(2)$$

17.$$\int_{0}^{\pi/\alpha}\sin x\ln\left[\tan\left(\frac{x}{2}\right)\right]\mathrm dx=\ln\left[\frac{1}{2}\sin\left(\frac{\pi}{\alpha}\right)\right]+\cos\left(\frac{\pi}{\alpha}\right)\ln\left[\cot\left(\frac{\pi}{2\alpha}\right)\right]$$

18.$$\int_{0}^{\pi/\alpha}\cos x\ln\left[\tan\left(\frac{x}{2}\right)\right]\mathrm dx=\sin\left(\frac{\pi}{\alpha}\right)\ln\left[\tan\left(\frac{\pi}{2\alpha}\right)\right]-\frac{\pi}{\alpha}$$

19.$$\int_{0}^{\pi}\tan x\ln\left[\cot\left(\frac{x}{2}\right)\right]\mathrm dx=\color{yellow}\frac{\pi^2}{4}$$

20.$$\int_{0}^{\pi/2}x^2\ln\left[\tan^2\left(\frac{x}{2}\right)\right]\mathrm dx=\color{purple}8\beta(4)-\pi^2C$$

21.$$\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx=\color{red}(-1)^n\left(\frac{1}{2}-\color{borwn}n\ln 2+\sum_{j=0}^{n-1}H_j^{*}\right)$$

22.$$\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{\ln^{n}(x)}{\sqrt{x^4-2x^2+1}}\mathrm dx=\color{gred}(-1)^n\Gamma(n+1)\eta(n)$$

23.$$\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{\ln^2(x)\ln(1+x)}{\sqrt{x^4-2x^2+1}}\mathrm dx=\color{blue}\zeta(2)-\color{green}\frac{5}{4}\zeta(5)$$

24.$$\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{1}{\sqrt{x^4-x^2+1}}\mathrm dx=\color{blue}\ln\left(\frac{3}{2}\right)$$

25.$$\int_{0}^{\infty}\frac{1-x}{1+x}\cdot \frac{\ln(1+x^n)}{\sqrt{x^4-2x^2+1}}\mathrm dx=\color{orange}-n\ln(2)$$

26.$$\int_{0}^{\infty}\frac{1-x}{1+x}\cdot \frac{\ln(x)}{\sqrt{x^4-2x^2+1}}\mathrm dx=\color{grey}-\ln(4)$$

27.$$\int_{0}^{\pi/2}\frac{\ln^2(\sin x)\ln(\cos x)}{\sin x \cos x}\mathrm dx=-\color{yellow}\frac{5}{2^5}\color{red}\zeta(4)$$

28.$$\int_{0}^{\pi/2}\frac{\ln(\sin x)\ln(\cos x)\ln(\sin x \cos x)}{\sin x \cos x}\mathrm dx=-\color{red}\frac{5}{2^4}\color{purple}\zeta(4)$$

29.$$\int_{0}^{\infty}\frac{x^k}{(1+\cosh x)^2}\mathrm dx=\color{red}2\Gamma(k+1)\color{purple}\cdot\frac{\eta(k)-\eta(k-2)}{3}$$

30.$$\int_{0}^{\infty}\frac{\tanh^2\left(\frac{x}{2}\right)}{xe^x}\frac{\mathrm dx}{\sinh x}=12\ln A-1-\left(\frac{1}{3}\ln 2+\ln \pi\right)+\frac{7}{2}\frac{\zeta(3)}{\pi^2}$$

31.$$\int_{0}^{\infty}\frac{\tanh^2\left(\frac{x}{2}\right)}{xe^x}\mathrm dx=12\ln A-1+2\left(\frac{1}{3}\ln 2-\ln \pi\right)$$

32.$$\int_{0}^{\infty}\frac{x}{e^{2\pi x}-1}\ln(x^2+x^4)\mathrm dx=\color{red}\frac{2\ln(2\pi)-3}{4}$$

33.$$\int_{0}^{\infty}\frac{\sqrt{x}}{e^{\pi/4 x}-1}\ln(x)\mathrm dx=\color{blue}\frac{\pi}{4}\cdot\color{green}\left[\zeta^{'}\left(\frac{3}{2}\right)-\color{brown}(\gamma+\ln\pi-2)\zeta\left(\frac{3}{2}\right)\right]$$

34.$$\int_{0}^{\pi/2}\sin(2x)\tan^2\left(\frac{x}{2}\right)\frac{\ln\cos\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)}\mathrm dx=\ln 4-3\ln^2(2)$$

35.$$\int_{0}^{\infty}\frac{\arccot(\alpha\sqrt{x})}{1+x}\mathrm dx=\pi\color{blue}\ln\left(1+\frac{1}{\alpha}\right)$$

36.$$\int_{0}^{\infty}\frac{\arccot(\alpha\sqrt{x})}{(1+x)^2}\mathrm dx=\color{blue}\frac{\pi}{2}\cdot\color{green}\frac{1}{1+\alpha}$$

37.$$\int_{0}^{\infty}\frac{\arccot(\sqrt{x}+\sqrt{1+x})}{1+x}\mathrm dx=\frac{\pi}{2}\color{blue}\ln 2$$

38.$$\int_{0}^{\infty}\frac{\arccot(\sqrt{x}+\sqrt{1+x})}{(1+x)^2}\mathrm dx=\color{grey}\frac{\pi}{8}$$

39.$$\int_{0}^{\infty}\frac{\arccot(\sqrt{x}-\sqrt{1+x})}{(1+x)^2}\mathrm dx=-\color{purple}\frac{3\pi}{8}$$

40.$$\int_{0}^{\infty}\frac{\arccot(\sqrt{1+x}+\sqrt{2+x})}{1+x}\mathrm dx=\color{red}C$$

41.$$\int_{0}^{\infty}\frac{\arccot(\sqrt{1+x}+\sqrt{2+x})}{(1+x)^3}\mathrm dx=\color{brown}\frac{1}{3!}$$

42.$$\int_{0}^{\infty}\frac{\arccot(\sqrt{x})\ln x}{1+x}\mathrm dx=\color{purple}\frac{\pi^3}{12}$$

43.$$\int_{0}^{\infty}\frac{\arccot(\sqrt{x})\ln x}{(1+x)^2}\mathrm dx=\color{purple}\pi\left(\ln 2-\color{red}\frac{1}{2}\right)$$

44.$$\int_{0}^{\infty}\frac{\arccot(\alpha\sqrt{x})\ln x}{(1+x)^2}\mathrm dx=\color{green}\pi\ln\left(\frac{\alpha}{\alpha+1}\cdot\alpha^{\frac{1}{\alpha^2-1}}\right)$$

45.$$\int_{0}^{\infty}\frac{\arccot^2(\sqrt{x}+\sqrt{1+x})}{(1+x)^2}\mathrm dx=\frac{1}{32}\color{red}(\pi-2)(\pi+2)$$

46.$$\int_{0}^{\infty}\frac{\arccot^2(\sqrt{x}-\sqrt{1+x})}{(1+x)^2}\mathrm dx=\frac{1}{32}\color{blue}(5\pi^2-4)$$

47.$$\int_{0}^{\infty}\frac{\arccot(\phi\sqrt{x})\ln x}{(1+x)^2}\mathrm dx=\color{green}\frac{\pi}{\phi^2}\color{red}\ln\frac{1}{\phi}$$

48.$$\int_{0}^{\infty}\frac{\arccot^2(\sqrt{1+x}+\sqrt{2+x})}{(1+x)^3}\mathrm dx=\color{blue}\frac{1!+\pi}{4!}-\color{red}\frac{\ln 2!}{3!}$$

49.$$\int_{0}^{\pi/4}\sqrt{\ln\left(\frac{1}{2}\csc^2(x)\right)}\mathrm dx=\frac{1}{2}\color{purple}\sqrt{\pi}$$

50.$$\int_{0}^{\pi/4}\cos\left(\frac{x}{2}\right)\ln\left(2\sin^2(x)\right)\mathrm dx=\color{green}2\sin\left(\frac{\pi}{2\alpha}\right)\color{red}\left[\ln(\left(1-\color{blue}\cos\left(\frac{\pi}{\alpha}\right)\right)-2\right]$$

51.$$\int_{0}^{\pi/4}\cos x\sqrt{\ln\csc(x)}\mathrm dx=\color{green}\frac{\sqrt{\pi}\color{red}\operatorname{erfc}\left(\frac{1}{2}\ln 2\right)-\color{purple}\sqrt{\ln 2}}{2}$$

52.$$\int_{0}^{\pi/4}\cos x\ln^{n}(2\sin^2 x)\mathrm dx=(-1)^n2^{n-1}\sqrt{2}n!$$

53.$$\int_{0}^{\infty}\ln\left(\frac{1}{x}\right)\cdot\frac{\mathrm dx}{1+x^{2^n}}=\color{red}\frac{\pi^2}{2^{2n}}\color{blue}\cot\left(\frac{\pi}{2^n}\right)\color{green}\csc\left(\frac{\pi}{2^n}\right)$$

54.$$\int_{0}^{\infty}\ln\left(\frac{1}{x}\right)\cdot\frac{\mathrm dx}{(1+x^{2^n})^2}=\color{red}\frac{\pi}{2^{3n}}\left[\color{blue}2^n+(2^n-1)\pi \cot\left(\frac{\pi}{2^n}\right)\right]\color{green}\csc\left(\frac{\pi}{2^n}\right)$$

55.$$\int_{0}^{\infty}x^{\alpha}\ln\left(\frac{1}{x}\right)\cdot\frac{\mathrm dx}{1+x^{2^n}}=\color{red}\frac{\pi^2}{2^{2n}}\color{blue}\cot\left(\frac{(\alpha+1)\pi}{2^n}\right)\color{green}\csc\left(\frac{(\alpha+1)\pi}{2^n}\right)$$

54.$$\int_{0}^{\infty}x^{\alpha}\ln\left(\frac{1}{x}\right)\cdot\frac{\mathrm dx}{(1+x^{2^n})^2}=\color{red}\frac{\pi}{2^{3n}}\left[\color{blue}2^n+(2^n-1-\alpha)\pi \cot\left(\frac{(\alpha+1)\pi}{2^n}\right)\right]\color{green}\csc\left(\frac{(\alpha+1)\pi}{2^n}\right)$$

55.$$\int_{0}^{\pi/2}\frac{\ln^2(\sin x)\ln^2(\cos x)}{\sin x \cos x}\ln(\sin x \cos x)\mathrm dx=\color{blue}\frac{2[2\zeta(5)-\zeta(2)\zeta(3)]}{\pi^2-1}$$

56.$$\int_{0}^{\pi/2}\frac{\ln(\sin x+\cos x)}{\sin x\cos x}\mathrm dx=\color{green}\lambda(2)$$

57.$$\int_{0}^{\pi/2}\frac{\ln(\sin x+\cos x)}{\sin x\cos x}\cdot x\mathrm dx=\color{red}\beta(3)$$

58.$$\int_{0}^{\infty}\frac{\sin(\alpha x)-\sin(\beta x)}{\sinh(\gamma x)}\mathrm dx=\color{green}\frac{\pi}{2\gamma}\cdot \left[\color{blue}\tanh\left(\frac{\pi}{2\gamma}\alpha\right)-\color{red}\tanh\left(\frac{\pi}{2\gamma}\beta\right)\right]$$

59.$$\int_{0}^{\infty}\frac{\cos(\alpha x)-\cos(\beta x)}{\cosh(\gamma x)}\mathrm dx=\color{green}\frac{\pi}{2\gamma}\cdot \left[\color{brown}\operatorname{sech}\left(\frac{\pi}{2\gamma}\alpha\right)-\color{purple}\operatorname{sech}\left(\frac{\pi}{2\gamma}\beta\right)\right]$$

60.$$\int_{0}^{1}\frac{\ln(x)\ln\left(\frac{\sqrt{x}+\sqrt{\alpha+x}}{x^{\beta}}\right)}{\sqrt{16x}}\mathrm dx=2\color{red}\sqrt{1+\alpha}-\color{blue}\sqrt{\alpha}-4\beta-\ln(1+\sqrt{1+\alpha})-\sqrt{\alpha}\ln\left(\frac{\alpha+\sqrt{\alpha(1+\alpha)}}{2}\right)$$

61.$$\int_{0}^{\pi/\alpha}\sin x\ln(\beta^2+2\beta \cos x+1)\mathrm dx=G(\alpha,\beta)$$

$$G(\alpha,\beta)=\color{red}\cos\left(\frac{\pi}{\alpha}\right)-1+\color{blue}\frac{(1+\beta)^2\ln[(1+\beta)^2]}{2\beta}-\color{green}\frac{[\beta^2+2\beta\cos\left(\frac{\pi}{\alpha}\right)+1]\color{brown}\ln[\beta^2+2\beta\cos\left(\frac{\pi}{\alpha}\right)+1]}{2\beta}$$

62.$$\int_{0}^{\pi}\ln\left[\beta^{2\alpha}\pm2\beta^{\alpha}\cos(\alpha x)+1\right]\mathrm dx=2\color{red}\alpha\color{blue}\pi\color{green}\ln\beta$$

63.$$\int_{0}^{\pi}x\ln\left[\beta^{2\alpha}\pm2\beta^{\alpha}\cos(\alpha x)+1\right]\mathrm dx=\color{red}\alpha\color{blue}\pi^2\color{green}\ln\beta$$

64.$$\int_{0}^{\pi}\left(\frac{\sin x}{\alpha^2+2\alpha\cos x+1}\right)^2\mathrm dx=\color{red}\frac{\pi}{2}\cdot\color{blue}\frac{1}{\alpha^2(\alpha^2-1)}$$

65.$$\int_{0}^{\pi}\left(\frac{\sin x}{\alpha^4+2\alpha^2\cos(2x)+1}\right)^2\mathrm dx=\color{red}\frac{\pi}{2}\cdot\color{blue}\frac{1}{(\alpha^4-1)(\alpha^2-1)^2}$$

66.$$\int_{0}^{\pi/2}\left(\frac{\sin x}{\alpha^4+2\alpha^2\cos(2x)+1}\right)^2\mathrm dx=\color{red}\frac{\pi}{4}\cdot\color{blue}\frac{1}{(\alpha^4-1)(\alpha^2-1)^2}$$

67.$$\int_{0}^{\pi/2}\left(\frac{\cos x}{\alpha^4+2\alpha^2\cos(2x)+1}\right)^2\mathrm dx=\color{red}\frac{\pi}{4}\cdot\color{blue}\frac{1}{(\alpha^4-1)(\alpha^2+1)^2}$$

68.$$\int_{0}^{\pi/2}\left(\frac{2\alpha \sin x}{\alpha^2+2\alpha\cos x+1}\right)^2\mathrm dx=\color{red}\frac{2\alpha}{1+\alpha^2}-\color{green}\frac{\pi}{2}+2\color{blue}\cdot\frac{\alpha^2+1}{\alpha^2-1}\arctan\left(\frac{\alpha-1}{\alpha+1}\right)$$

69.$$\int_{0}^{\pi/4}\left(\frac{2\sin x}{\alpha^4+2\alpha^2\cos(2x)+1}\right)^2\mathrm dx=\color{red}\frac{1-\alpha^4+2(1+\alpha^4)\color{blue}\arctan\left(\frac{\alpha^2-1}{\alpha^2+1}\right)}{\color{brown}(\alpha^8-1)(\alpha^2-1)^2}$$

70.$$\int_{0}^{\pi/4}\left(\frac{2\cos x}{\alpha^2+2\alpha^2\cos(2x)+1}\right)^2\mathrm dx=\color{red}\frac{1-\alpha^2+2(1+\alpha^2)\color{blue}\arctan\left(\frac{\alpha-1}{\alpha+1}\right)}{\color{brown}(\alpha^4-1)(\alpha+1)^2}$$

71.$$\int_{0}^{\pi/4}\left(\frac{2\cos x}{\alpha^4+2\alpha^2\cos(2x)+1}\right)^2\mathrm dx=\color{red}\frac{1-\alpha^4+2(1+\alpha^4)\color{blue}\arctan\left(\frac{\alpha^2-1}{\alpha^2+1}\right)}{\color{brown}(\alpha^8-1)(\alpha^2+1)^2}$$

72.$$\int_{0}^{\infty}\ln\left(x+\frac{1}{x}\right)\frac{\mathrm dx}{(1+x)^2}=\color{red}\frac{\pi}{2}\color{green}\ln 2$$

73.$$\int_{0}^{\infty}\ln\left(x+\frac{1}{x}\right)\frac{\mathrm dx}{(1+x)^3}=\color{red}\sqrt{\frac{3}{17}}\cdot\color{blue}\pi\ln 2$$

74.$$\int_{0}^{\infty}\ln\left(\frac{1+x^{\alpha}}{1+x^{\beta}}\right)\cdot \frac{1}{\ln x}\cdot\frac{\mathrm dx}{(1+x)^2}=\color{blue}\frac{\alpha-\beta}{2}$$

75.$$\int_{0}^{\infty}\ln\left(\frac{1+x^{\alpha}}{1+x^{\beta}}\right)\cdot \frac{x}{\ln x}\cdot\frac{\mathrm dx}{(1+x)^2(1+x^2)}=\color{blue}\frac{\alpha-\beta}{8}\color{purple}(\pi-2)$$

76.$$\int_{0}^{\infty}\frac{1}{\ln x}\cdot\ln\left(\frac{1+x^{\alpha}}{1+x^{\beta}}\right)\cdot \left(\frac{1-x}{1+x}\right)^2\cdot\frac{\mathrm dx}{(1+x^2)}=-\color{blue}\frac{\alpha-\beta}{4}\color{purple}(\pi-4)$$

77.$$\int_{0}^{\infty}\frac{\ln x}{(\alpha^2+x)^n}\mathrm dx=\color{red}\frac{H_{n-2}-2}{n-1}\cdot\color{blue}\frac{1}{\alpha^{2(n-1)}}$$

78.$$\int_{0}^{\infty}x^{\beta-1}\frac{\ln x}{(\alpha^2+x^{\beta})^n}\mathrm dx=\color{purple}\frac{H_{n-2}-2}{\beta^2(n-1)}\cdot\color{green}\frac{1}{\alpha^{2(n-1)}}$$

79.$$\int_{0}^{\infty}\frac{\ln^2(x)}{(x+e^2)^3}\mathrm dx=\color{red}\frac{\zeta(2)}{\color{blue}e^4}$$

80.$$\int_{0}^{\alpha \pi}\frac{x\cos^2(x)}{\cos^6(x)+\sin^6(x)}\mathrm dx=\frac{(\color{blue}\alpha \color{red}\pi)^2}{2}$$, $$\alpha\ge1$$

81.$$\int_{0}^{1}x^{\alpha-1}\arctan\left(\frac{1-x^{\alpha}}{1+x^{\alpha}}\right)\ln\left(\frac{1}{x}\right)\mathrm dx=\frac{\eta(2)+2\eta(1)}{(2\alpha)^2}$$

82.$$\int_{0}^{1}\ln\left(\frac{\alpha-x^2}{\alpha+x^2}\right)\frac{\mathrm dx}{x^2\sqrt{1-x^2}}=\color{blue}\left(\frac{\sqrt{\alpha-1}-\sqrt{\alpha+1}}{\sqrt{\alpha}}\right)\color{red}\pi$$

83.$$\int_{0}^{1}\left(\alpha+\ln\left(\frac{x}{1-x}\right)\right)^3\mathrm dx=\color{red}\alpha(\color{blue}\pi^2+\alpha^2)$$

84.$$\int_{0}^{1}\left(\alpha+\frac{\ln x}{1-x}\right)^2\ln\left(\frac{1}{x}\right)\mathrm dx=2(-1)^{\alpha}(2\alpha-3)\zeta(3)+\alpha^2$$

85.$$\int_{0}^{1}\left(1+\frac{\ln x}{1-x}\right)^3\ln x\mathrm dx=\color{red}12\zeta(4)-1^3$$

86.$$\int_{0}^{1}\left(2+\frac{\ln x}{1-x}\right)^3\ln x\mathrm dx=\color{blue}12\zeta(4)-2^3$$

87.$$\int_{0}^{1}\left(1+\ln\left(\frac{x}{1-x}\right)\right)^2\ln\left(\frac{1}{x}\right)\mathrm dx=\color{red}2\zeta(3)+1$$

88.$$\int_{0}^{\pi/2}\tan x \ln^2(\cos x)\ln(\sin x)\mathrm dx=-\color{blue}\frac{\zeta(4)}{2^3}$$

89.$$\int_{0}^{\pi/2}\tan x \ln(\cos x)\ln(\sin x)\ln(\sin x\cos x)\mathrm dx=-\color{green}\frac{5\zeta(4)}{2^5}$$

90.$$\int_{0}^{\pi/2}\tan x \ln(\cos x)\ln(\sin x)\ln(\tan x)\mathrm dx=\color{red}\frac{3\zeta(4)}{2^5}$$

91.$$\int_{0}^{\pi/2}\tan x \ln(\cos x)\ln^3(\sin x)\ln(\tan x)\mathrm dx=-\color{purple}\frac{3\zeta(6)}{2^8}$$

92.$$\int_{0}^{\pi/2}\arctan\left(\frac{4\sin^2 x}{3+2\cos^2 x}\right)\mathrm dx=\pi\arccot\left(\phi^3\right)$$

93.$$\int_{0}^{1}\frac{\ln x}{(2-x)^2}\mathrm dx=\color{green}\frac{1}{2}\ln 2$$

94.$$\int_{0}^{1}\frac{\ln x}{2-x}\mathrm dx=-\color{blue}\frac{\zeta(2)-\ln^2(2)}{2}$$

95.$$\int_{0}^{1}\left(\frac{\ln x}{2-x}\right)^2\mathrm dx=\color{red}\frac{\zeta(2)-\ln^2(2)}{2}$$

96.$$\int_{0}^{1}\frac{\ln^n(\alpha-x)}{\alpha-x}\mathrm dx=\color{blue}\frac{\ln^{n+1}(\alpha)-\ln^{n+1}(\alpha-1)}{n+1}$$

97.$$\int_{0}^{1}\frac{\ln^n(\alpha+x)}{\alpha+x}\mathrm dx=\color{blue}\frac{\ln^{n+1}(\alpha+1)-\ln^{n+1}(\alpha)}{n+1}$$

98.$$\int_{0}^{1}\frac{\arctan^2(x)}{x\sqrt{1-x^2}}\mathrm dx=\frac{\pi}{2}\operatorname{arcsinh}\left(\frac{4259}{9000}\right)$$

99.$$\int_{0}^{1}\left(\alpha+\frac{\ln^{\beta(x)}}{1-x}\right)^{\gamma}\mathrm dx=G(\alpha,\beta,\gamma)$$

100.$$\int_{0}^{1}\frac{x^{\alpha}\ln x}{x\sqrt{1-x^{\alpha}}}\mathrm dx=\color{blue}\left(\frac{2}{n}\right)^2\color{red}\left(\ln 2-1\right)$$

1.$$\int_{0}^{1}\frac{\operatorname{li}(x)\ln(-\ln x)}{x}\mathrm dx=1+\color{blue}\gamma$$

2.$$\int_{0}^{1}\frac{\operatorname{li}(x)\ln^{n}(x)\ln(-\ln x)}{x}\mathrm dx=\color{green}\frac{1}{(n+1)^2}\left[(n+1)!\color{blue}\gamma+G(n)\right]$$

Where $$G(1)=1, G(2)=-7, G(3)=38, G(4)=-226,...$$

3.$$\int_{0}^{1}\frac{\ln x\ln(1+x+x^2)}{x}\mathrm dx=-\color{blue}2\left(\frac{2}{3}\right)^2\color{red}\zeta(3)$$

4.$$\int_{0}^{1}\frac{\ln x\ln(1-x+x^2)}{x}\mathrm dx=\color{green}\frac{2}{3}\color{purple}\zeta(3)$$

5.$$\int_{0}^{1}\frac{\ln x\ln(1+x^{2^n})}{x}\mathrm dx=-\color{black}\frac{3}{2^{2n+2}}\color{purple}\zeta(3)$$

6.$$\int_{0}^{1}\frac{\ln^{2\alpha +1}x\ln(1+x^{2^n})}{x}\mathrm dx=-\color{black}\frac{G(n)}{2^{2^{\alpha}n+\alpha+1}}\color{purple}\zeta(2\alpha+3)$$, $$\alpha\ge0$$

7.$$\int_{0}^{\infty}\frac{\ln(1+\cosh x)}{1+\cosh x}\mathrm dx=2-\ln 2$$

8.$$\int_{0}^{\infty}\frac{\ln(1+\cosh x)}{1+\cosh x}\cdot x\mathrm dx=\color{brown}\ln 16$$

9.$$\int_{0}^{\infty}\frac{\ln(1+\cosh x)}{1+\cosh x}\cdot e^{-x}\mathrm dx=\color{red}\zeta(2)-2+\ln 2$$

10.$$\int_{0}^{\infty}\frac{\ln(1+\cosh x)}{1+\cosh x}\cdot (1+e^{-x})\mathrm dx=\color{purple}\zeta(2)$$

11.$$\int_{0}^{\infty}\frac{\ln(\cosh x)}{1+\cosh x}\cdot (1+e^{-x})\mathrm dx=\color{purple}\frac{(\pi-\ln 4)\color{green}(\pi+\ln 4)}{8}$$

12.$$\int_{-1}^{1}\frac{\ln(4-3x^2)}{2+x\sqrt{3}}\cdot \frac{\mathrm dx}{\sqrt{1-x^2}}=\color{blue}\pi\ln\left(\frac{16}{9}\right)$$

13.$$\int_{-1}^{1}\frac{\ln(4-3x^2)}{2+x\sqrt{3}}\cdot \frac{\mathrm dx}{\sqrt{1+x^2}}=\color{red}\pi\ln\left(\sqrt{\frac{15}{7}}\right)$$

14.$$\int_{-1}^{1}\frac{\mathrm dx}{\sqrt{1-x^2}(\alpha-\beta x^2)}=\color{red}\frac{\pi}{2}\cdot\frac{1}{\sqrt{\alpha(\alpha-\beta)}}$$

15.$$\int_{-1}^{1}\frac{x\ln\left(\frac{2-x\sqrt{3}}{2+x\sqrt{3}}\right)}{4-3x^2}\cdot \frac{\mathrm dx}{\sqrt{1-x^2}}=\color{red}\frac{\pi\ln(4)}{\sqrt{3}}$$

16.$$\int_{-1}^{1}\frac{x^3\ln\left(\frac{2-x\sqrt{3}}{2+x\sqrt{3}}\right)}{4-3x^2}\cdot \frac{\mathrm dx}{\sqrt{1-x^2}}=\color{blue}\frac{14\pi\ln(4)}{9\sqrt{3}}$$

17.$$\int_{-\infty}^{\infty}\frac{\ln\cosh x}{1+\cosh x}\mathrm dx=\color{red}\pi-\color{purple}\ln 4$$

18.$$\int_{0}^{\infty}e^{-2x}\frac{\ln\cosh x}{1+\cosh x}\mathrm dx=-\color{green}\left(\frac{\pi}{2}-1-\ln 2\right)\color{red}\left(\frac{\pi}{4}-2+\ln 2\right)$$

19.$$\int_{0}^{1}\frac{\ln(1+\alpha x)}{\sqrt{(1+\alpha x)^3}}\cdot\frac{\mathrm dx}{\sqrt{1-x}}=-\color{blue}\frac{4}{\alpha(\alpha+1)}\left(\alpha-\sqrt{\alpha}\color{red}\arccos\left(\frac{1}{\sqrt{\alpha+1}}\right)\right)$$

20.$$\int_{0}^{\pi}\frac{(9-4\cos x)\cos x\cos\left(\frac{x}{2}\right)}{\sqrt{1+\sin\left(\frac{x}{2}\right)}}\mathrm dx=\color{red}2\sqrt{2}$$

21.$$\int_{0}^{\pi/2}\frac{(9-4\cos x)\cos x\cos\left(\frac{x}{2}\right)}{\sqrt{1+\sin\left(\frac{x}{2}\right)}}\mathrm dx=\color{brown}2\sqrt{3}$$

22.$$\int_{0}^{\pi/3}\frac{(9-4\cos x)\cos x\cos\left(\frac{x}{2}\right)}{\sqrt{1+\sin\left(\frac{x}{2}\right)}}\mathrm dx=\color{orange}2\sqrt{5}$$

23.$$\int_{0}^{2\pi/3}\frac{(9-4\cos x)\cos x\cos\left(\frac{x}{2}\right)}{\sqrt{1+\sin\left(\frac{x}{2}\right)}}\mathrm dx=\color{green}\sqrt{21}$$

24.$$\int_{0}^{\pi}\frac{(9-4\cos x)\sin x\cos\left(\frac{x}{2}\right)}{\sqrt{1+\sin\left(\frac{x}{2}\right)}}\mathrm dx=\color{green}6+\color{blue}\pi$$

25.$$\int_{0}^{\pi}\frac{(9-4\cos x)\cos x\sin\left(\frac{x}{2}\right)}{\sqrt{1+\sin\left(\frac{x}{2}\right)}}\mathrm dx=\color{green}-2-\color{purple}\frac{3}{2}\pi$$

26.$$\int_{0}^{1}\frac{4x^3-7x^2+3x}{\sin(\pi x)}\mathrm dx=7\color{blue}\frac{\zeta(3)}{\pi^3}$$

27.$$\int_{0}^{1}\frac{x^4-3x^2+2x}{\sin(\pi x)}\mathrm dx=93\color{green}\frac{\zeta(5)}{\pi^5}$$

28.$$\int_{0}^{1}\frac{x}{\sin(\pi x)}\sum_{k=0}^{n}{n\choose k}(-x)^k \mathrm dx=G(n)$$, $$n\ge1$$

29.$$\int_{0}^{1}\frac{x\ln(\alpha-x^2)}{\sqrt{1-x^2}}\mathrm dx=\ln(\alpha)+\color{purple}\sqrt{\alpha-1}\color{blue}\arctan\left(2\frac{\sqrt{\alpha-1}}{\alpha-2}\right)-\color{red}2$$

30.$$\int_{0}^{1}\frac{x}{\sqrt{1-x^2}}\ln\left(\frac{\alpha+\sqrt{1+\alpha^2}}{\alpha-\sqrt{1+\alpha^2}}\right)\mathrm dx=\color{orange}\frac{\alpha \pi}{2}+\color{grey}\ln\left(\frac{\alpha+1}{\alpha-1}\right)-\sqrt{\alpha^2-2}\color{red}\left[\pi-\arctan(\alpha\sqrt{\alpha^2-2})\right]$$

31.$$\int_{0}^1{}\frac{(1+x)^n}{\ln^2(x)+\pi^2}\mathrm dx=\frac{1}{\pi}\color{blue}\sum_{k=0}^{n}(-1)^k{n \choose k}\operatorname{Si}((k+1)\pi)$$

32.$$\int_{0}^{1}\frac{(1+x)^n}{\ln^2(x)+\pi^2}\ln x\mathrm dx=-\color{green}\sum_{k=0}^{n}(-1)^k{n \choose k}\operatorname{Ci}((k+1)\pi)$$

33.$$\int_{0}^{\pi}\frac{\ln\left[1+\sin\left(\frac{x}{2}\right)\right]}{1+\sin\left(\frac{x}{2}\right)}\cdot \cos\left(\frac{x}{2}\right)\mathrm dx=\ln^2(2)$$

34.$$\int_{0}^{\pi}\frac{\ln^n\left[\sin\left(\frac{x}{2}\right)\right]}{1+\sin\left(\frac{x}{2}\right)}\cdot \cos\left(\frac{x}{2}\right)\mathrm dx=(-1)^nG(n)\zeta(n+1)$$

35.$$\int_{0}^{\pi}\frac{\ln\left[\sin\left(\frac{x}{2}\right)\right]}{1+\sin\left(\frac{x}{2}\right)}\cdot \cos\left(\frac{x}{2}\right)\mathrm dx=-\color{purple}\zeta(2)$$

36.$$\int_{0}^{\pi}\frac{\ln^n\left[\sin\left(\frac{x}{2}\right)\right]}{1+\sin\left(\frac{x}{2}\right)}\cdot \cos^3\left(\frac{x}{2}\right)\mathrm dx=\color{blue}(-1)^nG(n)$$

$$G(1)=-\frac{3}{2},G(2)=\frac{7}{2}, G(3)=-\frac{45}{4}$$

37.$$\int_{0}^{\pi}\frac{\cos^3\left(\frac{x}{2}\right)}\cdot\frac{\mathrm dx}=-\color{brown}\ln 4$$

38.$$\int_{0}^{\pi}\frac{\cos^5\left(\frac{x}{2}\right)}\cdot\frac{\mathrm dx}=-\ln\left(\frac{9}{4}\right)$$

39.$$\int_{0}^{\pi}\frac{\cos^7\left(\frac{x}{2}\right)}\cdot\frac{\mathrm dx}=-\ln\left(\frac{729}{400}\right)$$

38.$$\int_{0}^{\pi}\frac{\cos^3\left(\frac{x}{2}\right)}\cdot\frac{\sin^n\left(\frac{x}{2}\right)}\mathrm dx=-2\ln\left(\color{red}\frac{n+2}{n+1}\right)$$

39.$$\int_{0}^{\pi}\frac{\cos^5\left(\frac{x}{2}\right)}\cdot\frac{\sin^n\left(\frac{x}{2}\right)}\mathrm dx=-2\color{purple}\ln\left[\color{red}\frac{(n+2)(n+3)}{(n+1)(n+4)}\right]$$

40.$$\int_{0}^{1}\arctan^n\left(\sqrt{\frac{x^2}{\alpha-x^2}}\right)\mathrm dx=\color{red}G(\alpha, n)$$

41.$$\int_{0}^{\infty}xe^{-\frac{x}{\alpha}x}\ln x\cdot\frac{\mathrm dx}{\sinh\left(\frac{\pi}{\alpha}x\right)}=\color{blue}\frac{\alpha^2}{12}\left[1+\ln\left(\frac{\alpha}{A^{12}}\right)\right]$$

42.$$\int_{0}^{\infty}xe^{-\frac{x}{\alpha}x}\ln x\cdot\frac{\mathrm dx}{\cosh\left(\frac{\pi}{\alpha}x\right)}=\color{purple}\frac{\alpha^2}{24}\left[1+\ln\left(\frac{2\alpha}{A^{12}}\right)\right]$$

43.$$\int_{0}^{\infty}x\ln x\cdot\frac{\mathrm dx}{\sinh\left(\frac{\pi}{\alpha}x\right)}=\color{green}\frac{\alpha^2}{12}\left[3+\ln\left(\frac{16\alpha^3}{A^{36}}\right)\right]$$

44.$$\int_{0}^{1/\alpha}\frac{\ln x \ln(1+\alpha x)}{x}\mathrm dx=\color{purple}-\eta(3)-\color{brown}\eta(2)\ln \alpha$$

45.$$\int_{0}^{1/\alpha}\frac{\ln x \ln(1-\alpha x)}{x}\mathrm dx=\color{red}\zeta(3)+\color{blue}\zeta(2)\ln \alpha$$

46.$$\int_{0}^{1/\alpha}\frac{\ln(1+\alpha x) \ln(1-\alpha x)}{x^2}\mathrm dx=\color{red}\alpha\eta(2)-\alpha\color{blue}\ln^2(2)$$

47.$$\int_{0}^{\infty}\frac{\ln^2(\alpha^2+x^2)}{\alpha^2+x^2}\mathrm dx=\frac{\pi}{\alpha}\color{brown}\left[\zeta(2)+2\ln^2(2\alpha)\right]$$

48.$$\int_{-1}^{1}\frac{x^2}{\pi^2+4\operatorname{arctanh^2}x}\mathrm dx=\frac{\color{red}2}{\color{green}3^2\times 5}$$

49.$$\int_{-1}^{1}\frac{x\ln(1-x)}{(\pi^2+4\operatorname{arctanh^2}x)^2}\mathrm dx=-\frac{\color{red}1}{\color{purple}2^4\times 3\times \color{brown}5}$$

50.$$\int_{-\infty}^{+\infty}\frac{x\sin(\alpha x)+\sin\left(\frac{x}{\alpha}\right)}{(1+x^2)^n}\mathrm dx=G(n,\alpha)$$, $$n\ge1$$

$$G(1,\alpha)=\frac{\pi}{e^{\alpha}}$$

$$G(2,\alpha)=\frac{\alpha}{1!\times 2}\frac{\pi}{e^{\alpha}}$$

$$G(3,\alpha)=\frac{\alpha(\alpha+1)}{2!\times 2^2}\frac{\pi}{e^{\alpha}}$$

$$G(4,\alpha)=\frac{\alpha(\alpha^2+3\alpha+3)}{3!\times 2^3}\frac{\pi}{e^{\alpha}}$$

51.$$\int_{-\infty}^{+\infty}\frac{\sin(\alpha x)+\sin\left(\frac{x}{\alpha}\right)}{1+x^2}x\mathrm dx=\frac{\pi}{e^{\alpha}}\color{blue}\left(1+e^{\frac{\alpha^2-1}{\alpha}}\right)$$

52.$$\int_{0}^{\infty}\frac{\sin x\sin\left(\frac{x}{2^n}\right)}{x^2}\cdot \prod_{j=0}^{n}\cos\left(\frac{x}{2^j}\right)\mathrm dx=\color{purple}\frac{1}{2^n}\cdot \color{green}\frac{\pi}{4}$$

53.$$\int_{0}^{\infty}\frac{\sin^2\left(\frac{x}{2^n}\right)}{x^2}\cdot \prod_{j=0}^{n}\cos\left(\frac{x}{2^j}\right)\mathrm dx=\color{purple}\frac{1}{2^{2n}}\cdot \color{blue}\frac{\pi}{4}$$

54.$$\int_{0}^{\infty}\frac{\sin^3\left(\frac{x}{2^n}\right)}{x^2}\cdot \prod_{j=0}^{n}\cos\left(\frac{x}{2^j}\right)\mathrm dx=\color{purple}G(n)$$

55.$$\int_{0}^{\infty}\frac{\sin x\sin\left({2^n}x\right)}{x^2}\cdot \prod_{j=0}^{n}\cos\left(\frac{x}{2^j}\right)\mathrm dx=\color{purple}\frac{\pi}{4}$$

56.$$\int_{0}^{\infty}\frac{\sin^3\left(x\right)}{x^2}\cdot \cos x\mathrm dx=\color{grey}\ln\left(2^{1/2}\right)$$

57.$$\int_{0}^{\infty}\frac{\sin^3\left(2x\right)}{x^2}\cdot \cos x\cos(2x)\mathrm dx=\color{red}\ln\left(\frac{3^{3/4}\cdot 7^{7/16}}{5^{5/8}}\right)$$

58.$$\int_{0}^{\infty}\frac{\sin^3\left({2^n}x\right)}{x^2}\cdot \prod_{j=0}^{n}\cos\left(\frac{x}{2^j}\right)\mathrm dx=\color{purple}G(n)$$

59.$$\int_{0}^{\infty}\left(\frac{\sin x}{x}\right)^3\cos x \ln x\mathrm dx=\color{red}\frac{3\pi(1-\ln 4)-2\pi\gamma}{8}$$

60.$$\int_{0}^{\infty}\frac{\sinh(\alpha x)}{\cosh^2(\alpha x)}\ln^n\cosh (\alpha x)\mathrm dx=\frac{\color{red}\Gamma(n+1)}{\alpha}$$

61.$$\int_{0}^{\infty}\frac{\sinh(\alpha x)}{\cosh^3(\alpha x)}\ln^n\cosh (\alpha x)\mathrm dx=\frac{\color{red}\Gamma(n+1)}{\color{brown}2^{n+1}\alpha}$$

62.$$\int_{0}^{\infty}\frac{\sinh(\alpha x)}{\cosh^3(\alpha x)}\ln^n\cosh (\alpha x)\mathrm dx=\frac{\color{red}\Gamma(n+1)}{\color{brown}3^{n+1}\alpha}$$

63.$$\int_{0}^{\infty}\frac{\sinh(\alpha x)}{\cosh^k(\alpha x)}\ln^n\cosh (\alpha x)\mathrm dx=\frac{\color{red}\Gamma(n+1)}{\color{brown}(k-1)^{n+1}\alpha}$$

64.$$\int_{0}^{\infty}\frac{\ln^2(x)}{Ai^2(x)+Bi^2(x)}\mathrm dx=\frac{146}{125}\color{blue}\pi$$

65.$$\int_{0}^{1}\frac{\ln x\ln^3(1-x)}{x(1-x)}\mathrm dx=6[3\zeta(5)-\zeta(2)\zeta(3)]$$

66.$$\int_{0}^{\infty}\frac{\cos(2^n x)-\cos(x)\cos(2x)\cos(4x)\times\cdots \times \cos(2^{n-1}x)}{x^2}\mathrm dx=-\color{blue}\frac{\pi}{2^{2-n}}$$, $$n\ge1$$

67.$$\int_{0}^{\infty}\frac{\cos^4(2^n x)-\cos(x)\cos(2x)\cos(4x)\times\cdots \times \cos(2^{n-1}x)}{x^2}\mathrm dx=-\color{red}\frac{\pi}{2^{1-n}}$$, $$n\ge1$$

68.$$\int_{0}^{\infty}\left[\frac{1}{1+x}-\frac{2}{(1+x)^2}+\frac{1}{(1+x)^3}\right]\cdot \left(\frac{\ln x}{x}\right)^2\mathrm dx=\color{red}\zeta(2)$$

69.$$\int_{0}^{\infty}\left(\frac{12+\sqrt{x}}{(1+x)^2}-\frac{12}{e^x}\right)\frac{\ln x}{x}\mathrm dx=\color{brown}\pi(\pi-1)-6\gamma^2$$

70.$$\int_{0}^{\infty}\left(\frac{12+\sqrt{x}}{(1+x)^3}-\frac{12}{e^x}\right)\frac{\ln x}{x}\mathrm dx=\color{green}6(1-\gamma^2)+\color{blue}\pi(\pi-1)$$

71.$$\int_{0}^{\infty}\left(\frac{12+\sqrt{x}}{(1+\sqrt{x})^2}-\frac{12}{e^{\alpha x}}\right)\frac{\ln x}{x}\mathrm dx=\color{brown}7\pi^2-6(\gamma+\ln \alpha)^2$$

72.$$\int_{0}^{\infty}\left(\frac{12+\sqrt{x^3}}{(1+x)^2}-\frac{12}{e^{\alpha x}}\right)\frac{\ln x}{x}\mathrm dx=\color{brown}\pi(\pi+1)-6(\gamma+\ln \alpha)^2$$

73.$$\int_{0}^{\infty}\left(\frac{12+\sqrt{x}}{(1+x)^2}-\frac{12}{e^{\alpha x}}\right)\frac{\ln x}{x}\mathrm dx=\color{brown}\pi(\pi-1)-6(\gamma+\ln \alpha)^2$$

74.$$\int_{0}^{\infty}\frac{x-\sqrt{x}}{x}\cdot \frac{\ln x}{(1+x)^2}\mathrm dx=\pi$$

75.$$\int_{0}^{\infty}\frac{x-\sqrt{x}}{x}\cdot \frac{\ln(1+x)}{(1+x)^2}\mathrm dx=\pi\left(\ln 2-\frac{1}{2}\right)-\color{green}1$$

76.$$\int_{0}^{1}\frac{1+3x}{1+x}\ln x\ln^3(1+x)\frac{\mathrm dx}{x}=3\zeta(2)\color{red}\zeta(3)-\color{blue}\frac{93}{16}\zeta(5)$$

77.$$\int_{0}^{1}\frac{1+3x}{1+x}\ln x\ln^2(1+x)\frac{\mathrm dx}{x}=-\color{red}\frac{\zeta(4)}{4}$$

78.$$\int_{0}^{1}\frac{1+3x}{1+x}\ln x\ln(1+x)\frac{\mathrm dx}{x}=-\color{purple}{\zeta(3)}$$

79.$$\int_{0}^{1}\frac{\ln(1-x)}{x^2-2x+2}\mathrm dx=-C$$

80.$$\int_{0}^{1}\frac{\ln^n(1-x)}{x^2-2x+2}\mathrm dx=(-1)^n\color{blue}\Gamma(n+1)\color{red}\beta(n+1)$$

81.$$\int_{0}^{1}\frac{\ln(1-x)}{2x^2-2x+1}\mathrm dx=\color{green}-C-\frac{\pi}{4}[\ln 2+0]$$

82.$$\int_{0}^{1}\frac{\ln(1-x)}{(2x^2-2x+1)^2}\mathrm dx=\color{red}-C-\frac{\pi}{4}[\ln 2+1]$$

83.$$\int_{0}^{\infty}\frac{x^n-\sin^n(x)}{x^{n+2}}\ln x\mathrm dx=\frac{a}{b}\pi\left(H_{n+1}-\gamma+\frac{c}{d}\ln n\right)$$

84.$$\int_{0}^{\infty}\frac{x-\sin(x)}{x^{3}}\ln x\mathrm dx=\color{purple}\frac{\pi}{4}(H_2-\color{black}\gamma)-\frac{\pi}{4}\ln 1$$

85.$$\int_{0}^{\infty}\frac{x^2-\sin^2(x)}{x^{4}}\ln x\mathrm dx=\color{blue}\frac{\pi}{3}(H_3-\color{brown}\gamma)-\frac{\pi}{3}\ln 2$$

86.$$\int_{0}^{\infty}\frac{x^4-\sin^4(x)}{x^{6}}\ln x\mathrm dx=\color{red}\frac{7\pi}{15}(H_5-\color{grey}\gamma)-\frac{\pi}{2}\ln 4$$

87.$$\int_{0}^{\infty}\frac{x^n-\sin^n(x)}{x^{n+1}}\sin x\ln x\mathrm dx=G(n)$$

88.$$\int_{0}^{\infty}\frac{x-\sin(x)}{x^{2}}\sin x\ln x\mathrm dx=\frac{\pi}{2}(\ln 2-1)$$

89.$$\int_{0}^{\infty}\frac{x^2-\sin^2(x)}{x^{3}}\sin x\ln x\mathrm dx=-\color{green}\frac{\pi}{4^2}[\color{blue}2\gamma+3^2(1-\color{red}\ln 3)]$$

90.$$\int_{0}^{\pi/2}\ln\left[\frac{(1+\sin x)^{\sin x}}{1+\cos x}\right]\mathrm dx=-2C-1+\frac{\pi}{2}(1+\ln 2)$$

91.$$\int_{0}^{\pi/2}\ln\left[\frac{(1+\sin x)^{1+\sin x}}{1+\cos x}\right]\mathrm dx=\frac{1}{2}(\pi-2)$$

92.$$\int_{0}^{\pi/2}\ln\left[\frac{(1+\sin x)^{1+\cos x}}{1+\cos x}\right]\mathrm dx=\ln 4-1$$

93.$$\int_{0}^{\pi/2}\ln\left[\frac{(1+\sin x)^{\cos x}}{1+\cos x}\right]\mathrm dx=-2C-1+\left(2+\frac{\pi}{2}\right)\ln 2$$

94.$$\int_{0}^{1}\ln\left(\frac{x-1+\sqrt{1+x^2}}{x+1+\sqrt{1+x^2}}\right)\mathrm dx=-2\ln(1+\sqrt{2})$$

95.$$\int_{0}^{1}\frac{1-x+\sqrt{1+x^2}}{x+1+\sqrt{1+x^2}}\cdot\ln\left(\frac{1-x+\sqrt{1+x^2}}{x+1+\sqrt{1+x^2}}\right)\mathrm dx=\color{red}\frac{1-\sqrt{2}}{4}-\color{blue}\frac{\ln(1+\sqrt{2})}{4}\left(2\sqrt{2}-3+\color{purple}\ln(1+\sqrt{2})\right)$$

96.$$\int_{0}^{\infty}\frac{e^{-\alpha/x}\sin\left(\frac{\beta}{x}\right)\ln(x)}{x}\mathrm dx=\color{red}\frac{\color{blue}2\gamma+\ln(\alpha^2+\beta^2)}{2}\cdot\arctan\left(\frac{\alpha}{\beta}\right)$$

97.$$\int_{0}^{\infty}\frac{e^{-\alpha/x}\sin\left(\frac{\beta}{x}\right)\ln^2(x)}{x}\mathrm dx=\color{red}\frac{\color{blue}3[2\gamma+\ln(\alpha^2+\beta^2)]^2+2\left[\pi^2-2\arctan\left(\frac{\alpha}{\beta}\right)\right]}{12}\cdot\arctan\left(\frac{\alpha}{\beta}\right)$$

98.$$\int_{0}^{\infty}\frac{e^{-\alpha/x}\sin^2\left(\frac{\beta}{x}\right)\ln(x)}{x}\mathrm dx=\frac{\ln(4\alpha^2+\beta^2)+4\gamma\ln\left(\frac{4\alpha^2+\beta^2}{\beta^2}\right)}{16}-\frac{\ln^2(\beta)+\arctan^2\left(\frac{2\alpha}{\beta}\right)}{4}$$

99.$$\int_{0}^{\infty}\sin(\alpha x)\arctan\left(\frac{\beta}{x}\right)\mathrm dx=\color{red}\frac{\pi}{2\alpha}\color{blue}\left(1-\frac{1}{e^{\alpha\beta}}\right)$$

100.$$\int_{\alpha}^{\beta}x\arccos\left(\frac{x}{\sqrt{(\alpha+\beta)x-\alpha\beta}}\right)\mathrm dx=\color{blue}\left(1-\frac{1}{4(\alpha+\beta)^2}\right)\cdot\color{red}\frac{\pi}{8}$$

- 1.$$\int_{-\pi/2}^{\pi/2}\cos^2 x\sin(\alpha+\beta\tan x)\mathrm dx=\frac{1+\beta}{2}\cdot\color{purple}\frac{\pi}{e^{\beta}}\color{red}\sin(\alpha)$$

2.$$\int_{\pi/2}^{\pi/2}\sin^2 x\sin(\alpha+\beta\tan x)\mathrm dx=\frac{1-\beta}{2}\cdot\color{purple}\frac{\pi}{e^{\beta}}\color{red}\sin(\alpha)$$

3.$$\int_{-\pi/2}^{\pi/2}\cos(2x)\cos(\alpha+\beta\tan x)\mathrm dx=\beta\cdot\color{purple}\frac{\pi}{e^{\beta}}\color{red}\cos(\alpha)$$

4.$$\int_{-\pi/2}^{\pi/2}\cos(2x)\sin(\alpha+\beta\tan x)\mathrm dx=\beta\cdot\color{purple}\frac{\pi}{e^{\beta}}\color{red}\sin(\alpha)$$

5.$$\int_{-\pi/2}^{\pi/2}\sin(2x)\cos(\alpha+\beta\tan x)\mathrm dx=-\beta\cdot\color{brown}\frac{\pi}{e^{\beta}}\color{black}\sin(\alpha)$$

6.$$\int_{-\pi/2}^{\pi/2}\sin(2x)\sin(\alpha+\beta\tan x)\mathrm dx=\beta\cdot\color{green}\frac{\pi}{e^{\beta}}\color{blue}\cos(\alpha)$$

7.$$\int_{0}^{\infty}\frac{x-\sin x}{x^3}\cdot \sin^{2n}(x)\mathrm dx=\frac{(2n-1)!!}{(2n)!!}\color{red}\cdot \frac{\pi}{4}$$

8.$$\int_{0}^{\infty}\frac{x-\sin x}{x^3}\cdot \ln x\mathrm dx=(3-2\gamma)\color{red}\cdot \frac{\pi}{8}$$

9.$$\int_{0}^{\infty}\frac{\sqrt{x}}{1+x}\ln\left(\frac{2\alpha\sqrt{x}}{1+x}\right)\mathrm dx=2-\pi+\color{red}\frac{\pi}{2}\ln\left(\frac{2}{a}\right)-\color{blue}2\ln \left(\frac{1}{a}\right)$$

10.$$\int_{0}^{\infty}\frac{x^3}{1+x}\ln\left(\frac{x^3}{1+x}\right)\ln(x)\mathrm dx=\frac{1}{6^3}\left[999\zeta(3)-66\zeta(2)-10(8\times 7+8\times 6\color{red}\ln2)\right]$$

11.$$\int_{0}^{1}\frac{\ln(1-x)\ln^2(x)}{\sqrt{x(1-x)}}\mathrm dx=2\pi\color{red}[\color{blue}\zeta(3)-4\color{green}\ln^3(2)]$$

12.$$\int_{0}^{1}\frac{x(x-2)+2\ln(1+x)}{x^2\sqrt{x(1-x)}}\mathrm dx=\frac{\color{red}1}{\color{green}3}(\color{brown}4\sqrt{\color{grey}2}-\color{black}5)\color{purple}\pi$$

13.$$\int_{0}^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha}\tan(x)\tan\left(\frac{x}{2}\right)\right]\sqrt{\sin(x)\tan\left(\frac{x}{2}\right)}\mathrm dx=-\color{red}\frac{\ln \alpha}{\sqrt{2}}$$

14.$$\int_{0}^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha}\sin(x)\sin\left(\frac{x}{2}\right)\right]\sqrt{\sin(x)\tan\left(\frac{x}{2}\right)}\mathrm dx=-\frac{3+\color{blue}\ln\left(\frac{\alpha^2}{2}\right)}{2\sqrt{2}}$$

15.$$\int_{0}^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha}\cos(x)\cos\left(\frac{x}{2}\right)\right]\sqrt{\sin(x)\tan\left(\frac{x}{2}\right)}\mathrm dx=-\color{green}\frac{3-\ln\left(\frac{\alpha^2}{2}\right)}{2\sqrt{2}}$$

16.$$\int_{0}^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha}\tan\left(\frac{x}{2}\right)\right]\sqrt{\sin(x)\tan\left(\frac{x}{2}\right)}\mathrm dx=-\color{brown}\frac{\ln (2\alpha)}{\sqrt{2}}$$

17.$$\int_{0}^{\pi/2}x\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha}\tan\left(\frac{x}{2}\right)\right]\sqrt{\sin(x)\tan\left(\frac{x}{2}\right)}\mathrm dx=\color{purple}\frac{\pi}{2}\cdot\frac{\ln 2-1}{\sqrt{2}}-\color{blue}\frac{\ln(\alpha^2)}{2\sqrt{2}}$$

18.$$\int_{0}^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left(\frac{1}{2}\sin x\right)\sqrt{\sin(x)\tan\left(\frac{x}{2}\right)}\mathrm dx=-\color{red}\frac{1}{\sqrt{2}}$$

19.$$\int_{0}^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha}\sin\left(\frac{x}{2}\right)\right]\sqrt{\sin(x)\tan\left(\frac{x}{2}\right)}\mathrm dx=-\color{orange}\frac{1+\ln(2\alpha^2)}{2\sqrt{2}}$$

20.$$\int_{0}^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha}\cos\left(\frac{x}{2}\right)\right]\sqrt{\sin(x)\tan\left(\frac{x}{2}\right)}\mathrm dx=-\color{red}\frac{1+\ln(\frac{\color{green}\alpha^2)}{2}}{2\sqrt{2}}$$

21.$$\int_{0}^{\pi/2}x\sqrt{\sin(x)\tan\left(\frac{x}{2}\right)}\mathrm dx=-\color{orange}4-\pi$$

22.$$\int_{0}^{\pi/3}\ln^4\left[\frac{\sin x}{\sin\left(x+\frac{\pi}{3}\right)}\right]=\color{red}\frac{\color{blue}1+2^{\color{red}4}}{3^{\color{green}5}}\cdot \color{blue}\pi^5$$

23.$$\int_{0}^{\pi/2}\ln^2\left[\alpha\tan\left(\frac{x}{2}\right)\right]\mathrm dx=\left(\frac{\pi}{2}\right)^3+\color{red}\frac{\pi}{2}\color{blue}\ln^2(\alpha)-\color{green}4C\color{blue}\ln(\alpha)$$

24.$$\int_{-\infty}^{+\infty}\frac{\mathrm dx}{(e^{-\alpha x}+\alpha x+1)^2+\pi^2}=\color{blue}\frac{1}{2\alpha}$$

25.$$\int_{-\pi/2}^{\pi/2}\sin(\alpha \tan x)\frac{\mathrm dx}{\tan x}=\pi\left(1-\frac{1}{e^{\alpha}}\right)$$

25. Where $$\alpha:=1,3,5,7,...$$

26.$$\int_{-\pi/2}^{\pi/2}\sin(\alpha+\beta\tan x)\frac{\mathrm dx}{\cot x}=\frac{\pi}{e^{\beta}}\cdot \cos(\alpha)$$

27.$$\int_{-\pi/2}^{\pi/2}\cos(\alpha+\beta\tan x)\frac{\mathrm dx}{\cot x}=-\frac{\pi}{e^{\beta}}\cdot \sin(\alpha)$$

28.$$\int_{0}^{1}\ln\left(\frac{\alpha^2+x}{\alpha^2-x}\right)\frac{\mathrm dx}{x\sqrt{x(1-x)}}=\color{blue}\frac{2\pi}{\alpha}\color{red}(\sqrt{\alpha^2+1}-\color{green}{\sqrt{\alpha^2-1}})$$

29.$$\int_{0}^{1}\ln x\ln\left(\frac{1-x}{1+x}\right)\frac{\mathrm dx}{1-x^2}=\frac{7}{8}\color{red}\zeta(3)$$

30.$$\int_{0}^{1}\left[\frac{\ln(1+x)}{1+x}\right]^2x\mathrm dx=-1+\color{red}\frac{\ln^3(2)}{3}+\color{blue}\frac{\ln^2(2)}{2}+\color{green}\frac{\ln(2)}{1}$$

31.$$\int_{0}^{1}\left[\frac{\ln(1+x)}{1+x}\right]^2\frac{\mathrm dx}{x}=\frac{\zeta(3)}{4}-1-\color{red}\frac{\ln^3(2)}{3}+\color{blue}\frac{\ln^2(2)}{2}+\color{green}\frac{\ln(2)}{1}$$

32.$$\int_{0}^{\infty}\left[\frac{\ln(1+x)}{1+x}\right]^n\mathrm dx=\color{green}\frac{n!}{(n-1)^{n+1}}$$

33.$$\int_{0}^{\pi/2}\sin x\arctan(\sin x)\mathrm dx=\color{purple}\frac{\pi}{2\alpha}(\sqrt{\alpha^2+1}-1)$$

34.$$\int_{0}^{\pi/2}\cos x\arctan(\sin x)\mathrm dx=\color{red}\arctan(\alpha)-\frac{\ln(\alpha^2+1)}{2\alpha}$$

35.$$\int_{0}^{\pi/2}\sin(2x)\arctan(\sin x)\mathrm dx=\color{green}\frac{\pi}{2}-1$$

36.$$\int_{0}^{\pi/2}\cos x\arctan\left(\frac{\alpha-\cos x}{\alpha+\cos x}\right)\mathrm dx=\color{red}\frac{\pi}{4}\color{green}(2\alpha+1-\sqrt{\alpha^2-1})$$

37.$$\int_{0}^{\pi/2}\sin x\arctan\left(\frac{\alpha-\cos x}{\alpha+\cos x}\right)\mathrm dx=\color{red}\frac{\alpha}{2}\ln\left(1+\frac{1}{\alpha^2}\right)+\arctan\left(\frac{\alpha-1}{\alpha+1}\right)$$

38.$$\int_{0}^{\pi/2}\sin x\arctan\left(\sqrt{\frac{\alpha-\cos x}{\alpha+\cos x}}\right)\mathrm dx=\color{red}\frac{\alpha}{2}-\frac{\sqrt{\alpha^2-1}}{2}+\arctan\sqrt{\frac{\alpha-1}{\alpha+1}}$$

39.$$\int_{0}^{\pi}\sin^{2n}(x)\arctan\left(\sqrt{\frac{\alpha-\cos x}{\alpha+\cos x}}\right)\mathrm dx=\color{red}\frac{(2n-1)!!}{(2n)!!}\cdot\left(\frac{\pi}{2}\right)^2$$

40.$$\int_{0}^{\pi}\sin^{2n-1}(x)\arctan\left(\sqrt{\frac{\alpha-\cos x}{\alpha+\cos x}}\right)\mathrm dx=\color{red}\frac{(2n-2)!!}{(2n-1)!!}\cdot\frac{\pi}{2}$$

41.$$\int_{0}^{\pi/4}\ln\cot(x)\arctan\sqrt{\frac{1-\cos x}{1+\cos x}}\cdot\frac{\mathrm dx}{x}=\frac{\color{purple}C}{2}$$

42.$$\int_{0}^{\pi/4}\ln^2[\tan(x)]\arctan\sqrt{\frac{1-\cos x}{1+\cos x}}\cdot\frac{\mathrm dx}{x}=\frac{1}{4}\color{red}\left(\frac{\pi}{2}\right)^3$$

43.$$\int_{0}^{\pi/4}x\ln\tan\left(x+\frac{\pi}{4}\right)\mathrm dx=\color{grey}\frac{7}{16}\color{red}\zeta(3)$$

44.$$\int_{0}^{\pi/4}\ln\tan\left(x+\frac{\pi}{4}\right)\mathrm dx=\color{blue}C$$

45.$$\int_{0}^{\pi/4}\ln\tan^2\left(x+\frac{\pi}{4}\right)\mathrm dx=\color{purple}\frac{\pi^3}{16}$$

46.$$\int_{0}^{\pi/4}\ln\left(1+\tan x\right)\mathrm dx=\frac{1}{8}\color{brown}\pi\ln 2$$

47.$$\int_{0}^{\pi/4}\ln\left(1-\tan x\right)\mathrm dx=\color{red}\frac{1}{8}(\pi\ln 2-8C)$$

48.$$\int_{0}^{\pi/4}x^2\ln\tan x\mathrm dx=\color{purple}\frac{\beta(4)}{2}-\color{green}\left(\frac{\pi}{4}\right)^2\color{red}C$$

49.$$\int_{0}^{\infty}\ln^{2n}(x)\ln(1+x)\frac{\mathrm dx}{x(1+x^2)}=\pi^{2n+2}\color{red}F(n)$$

50.$$\int_{0}^{\infty}\ln^{2n-1}(x)\ln(1+x)\frac{\mathrm dx}{x(1+x^2)}=\color{blue}G(n)$$

51.$$\int_{0}^{\infty}\ln^{2}(x)\ln(1+x)\frac{\mathrm dx}{x(1+x^2)}=\color{red}19\color{green}\frac{7^2\pi^4}{8!}$$

52.$$\int_{0}^{\infty}\ln^{2}(x)\ln(1+x^2)\frac{\mathrm dx}{x(1+x)}=\color{red}2\color{blue}\frac{(7\pi)^4}{8!}$$

53.$$\int_{0}^{\infty}\ln^{2n}(x)\ln(1+x^2)\frac{\mathrm dx}{x(1+x)}=\color{grey}\pi^{2n+2}F(n)$$

54.$$\int_{0}^{n\pi}\frac{x}{2+\alpha+\cos(2\alpha x)}\mathrm dx=\color{blue}\frac{(n\pi)^2}{2}\cdot\color{red}\frac{1}{\sqrt{(\alpha+1)(\alpha+3)}}$$

55.$$\int_{0}^{\infty}\frac{\ln^2(1+x^4)}{1+x^4}\mathrm dx=\int_{0}^{\infty}\frac{\ln^3(1+x^4)}{4x^4}\mathrm dx$$

56.$$\int_{0}^{\infty}\frac{\ln^{n}(1+x^4)}{1+x^4}\mathrm dx=\frac{3}{4(n+1)}\int_{0}^{\infty}\frac{\ln^{n+1}(1+x^4)}{x^4}\mathrm dx$$

57.$$\int_{0}^{1}\frac{\ln(x^2-x+1)}{x(x^-1)}\mathrm dx=\pi\color{green}\ln\left(\frac{4}{\pi}\right)$$

58.$$\int_{0}^{1}\sqrt{1-x}\ln(1-x+x^2)\mathrm dx=\frac{2}{3^2}\color{red}(3\pi-10)$$

59.$$\int_{0}^{1}\frac{\ln(1-x)\ln(1-x+x^2)}{1-x}\mathrm dx=\frac{2}{\color{green}3}\color{blue}\zeta(3)$$

60.$$\int_{0}^{1}\frac{\ln^3(1-x)\ln(1-x+x^2)}{1-x}\mathrm dx=2\left(\frac{5}{\color{green}3}\right)^2\color{blue}\zeta(5)$$

61.$$\int_{0}^{1}\frac{\ln^{n}(1-x)\ln(1-x+x^2)}{1-x}\mathrm dx=G(n)$$

62.$$\int_{0}^{1}\frac{\ln(1-x)\ln(1-2x+2x^2)}{1-x}\mathrm dx=\color{purple}\frac{35\zeta(3)+\pi^2\ln(4)}{32}$$

63.$$\int_{0}^{1}\frac{\ln 2-\ln(1+\sqrt{1-x^2})}{x^2}\mathrm dx=1-\color{purple}\ln 2$$

64.$$\int_{0}^{1}\frac{\ln 2-\ln(1+\sqrt{1+x^2})}{x^2}\mathrm dx=1-\color{purple}\sqrt{2}+\color{red}\ln\left(\color{green}\frac{1}{2}+\color{blue}\frac{1}{\sqrt{2}}\right)$$

65.$$\int_{0}^{1}\ln\left(\frac{1+\sqrt{1+x^2}}{1+\sqrt{1-x^2}}\right)\frac{\mathrm dx}{x^2}=\sqrt{2}-\color{red}\ln(1+\sqrt{2})$$

66.$$\int_{0}^{1}\ln x\ln\left(\frac{1+\sqrt{1+x^2}}{1+\sqrt{1-x^2}}\right)\frac{\mathrm dx}{x^2}=\frac{\pi}{2}-\color{blue}2\sqrt{2}+\color{red}2\ln(1+\sqrt{2})$$

67.$$\int_{0}^{1}\ln x\ln\left(\frac{1+\sqrt{1-x^2}}{2}\right)\frac{\mathrm dx}{x^2}=\frac{\pi}{2}-{2}+\color{red}\ln 2$$

68.$$\int_{0}^{1}\ln x\ln\left(\frac{1+\sqrt{1+x^2}}{2}\right)\frac{\mathrm dx}{x^2}=-2+2\sqrt{2}+\color{red}\ln 2-\color{orange}2\ln(1+\sqrt{2})$$

69.$$\int_{0}^{\infty}\frac{\ln x}{(1+x)^2(x+\alpha)}\mathrm dx=-\color{green}\frac{1}{2(\alpha-1)^2}\color{red}\ln^2(\alpha)$$

69. Where $$\alpha\ge-1$$

70.$$\int_{0}^{\infty}\frac{\ln^2(x)}{(1+x)^2(x+\alpha)}\mathrm dx=-\color{green}\frac{1}{3(\alpha-1)^2}[\color{red}\ln^3(\alpha)+\color{blue}\pi^2\ln(\alpha)+\alpha-1]$$

71.$$\int_{0}^{\infty}\frac{\ln^3(x)}{(1+x)^2(x+\alpha)}\mathrm dx=-\color{green}\frac{1}{4(\alpha-1)^2}\color{red}\ln^2(\alpha)[\color{blue}2\pi^2+\ln^2(\alpha)]$$

72.$$\int_{0}^{\infty}\frac{\ln^4(x)}{(1+x)^2(x+\alpha)}\mathrm dx=-\color{green}\frac{1}{15(\alpha-1)^2}[3\color{red}\ln^5(\alpha)+10\pi^2\color{blue}\ln^3(\alpha)+7\pi^4(\ln 2-(\alpha-1))]$$

73.$$\int_{0}^{\infty}\frac{\ln^5(x)}{(1+x)^2(x+\alpha)}\mathrm dx=-\color{green}\frac{1}{6(\alpha-1)^2}\color{red}\ln^2(\alpha)[\color{purple}7\pi^4+\ln^4(\alpha)+5\pi^2\ln^2(\alpha)]$$

73.$$\int_{0}^{\infty}\frac{\ln^{\alpha}(x)}{(1+x)^{\beta}(x+\gamma)}\mathrm dx=G(\alpha,\beta,\gamma)$$

74.$$\int_{0}^{\infty}\frac{\ln^2(1+x)}{(x+1)(x+2)}\mathrm dx=\frac{3}{2}\color{red}\zeta(3)$$

75.$$\int_{0}^{\infty}\frac{\ln^2(1+x)}{(x+1)^2(x+2)}\mathrm dx=2-\frac{3}{2}\color{red}\zeta(3)$$

76.$$\int_{0}^{\infty}\frac{\ln^2(2+x)}{(x+1)^3}\mathrm dx=2\color{green}\eta(1)-\color{red}\eta(2)$$

77.$$\int_{0}^{\infty}\frac{\ln^2(\alpha+x^2)}{\alpha+x^2}\mathrm dx=\frac{\pi^3+3\pi\ln^2(4\alpha)}{6\sqrt{\alpha}}$$

78.$$\int_{0}^{\infty}\frac{\ln^2\left(\frac{1}{4}+x^2\right)}{\frac{1}{4}+x^2}\mathrm dx=\frac{\pi^3}{3}$$

79.$$\int_{0}^{\infty}\frac{\ln^3\left(\frac{1}{4}+x^2\right)}{\frac{1}{4}+x^2}\mathrm dx=12\pi\zeta(3)$$

80.$$\int_{0}^{\infty}\frac{\ln^{2n-1}\left(\frac{1}{4}+x^2\right)}{\frac{1}{4}+x^2}\mathrm dx=G(n)$$

81.$$\int_{0}^{\infty}\frac{\ln\left(\frac{1}{4}+x+x^2\right)}{\left(\frac{1}{4}+x+x^2\right)^{\alpha}}\mathrm dx=\left(\frac{2^{\alpha}}{2\alpha-1}\right)^2\color{green}[(2\alpha-1)\ln(2)-1]$$

82.$$\int_{0}^{\infty}\frac{1}{4}\cdot\frac{\ln^2\left(\frac{1}{4}+x+x^2\right)}{\left(\frac{1}{4}+x+x^2\right)^{\alpha}}\mathrm dx=\frac{2^{2\alpha-1}}{(2\alpha-1)^3}\left(2+\color{red}(2\alpha-1)\ln 2[\color{green}(2\alpha-1)\ln 2-\color{purple}2]\right)$$

83.$$\int_{0}^{\pi/6}\frac{\ln\cos x}{1-\sin x}\mathrm dx=\frac{4+(3+\sqrt{3})\ln\frac{3}{4}}{2(1+\sqrt{3})}-\frac{\pi}{6}$$

84.$$\int_{0}^{\pi/3}\frac{\ln\cos x}{1-\sin x}\mathrm dx=1+\sqrt{3}-\color{red}(2+\sqrt{3})\ln 2-\color{green}\frac{\pi}{3}$$

85.$$\int_{0}^{\pi/2}\frac{(1+\beta\cos x)\cos x}{\alpha+\beta\cos x}\mathrm dx=1-\color{red}\frac{\alpha-1}{\beta}\cdot\frac{\pi}{2}+\color{green}\frac{2\alpha(\alpha-1)}{\beta\sqrt{(\alpha-\beta)(\alpha+\beta)}}\color{purple}\arctan\sqrt{\frac{\alpha-\beta}{\alpha+\beta}}$$

86.$$\int_{0}^{\pi/2}\frac{(\alpha-\beta+\beta\cos x)\cos x}{\alpha+\beta\cos x}\mathrm dx=1-\color{red}\frac{\pi}{2}+\color{orange}\frac{2\alpha}{\sqrt{(\alpha-\beta)(\alpha+\beta)}}\color{purple}\arctan\sqrt{\frac{\alpha-\beta}{\alpha+\beta}}$$

87.$$\int_{0}^{\pi/2}\frac{(\alpha-\beta+\beta\cos x)^2\cos(2x)}{\alpha+\beta\cos x}\mathrm dx=\frac{7}{3}\beta-\color{red}\alpha \pi+\color{orange}\frac{2(2\alpha^2-\beta^2)}{\sqrt{(\alpha-\beta)(\alpha+\beta)}}\color{purple}\arctan\sqrt{\frac{\alpha-\beta}{\alpha+\beta}}$$

88.$$\int_{0}^{1}\frac{\mathrm dx}{(x+\sqrt{1-x^2})^3}=\frac{1}{2}$$

89.$$\int_{0}^{1}\frac{\mathrm dx}{(x+\sqrt{1-x^2})^5}=\color{red}\frac{1}{3}$$

89.$$\int_{0}^{1}\frac{\ln x}{1+x-x^2}\mathrm dx=-\color{red}\left(\frac{\pi}{5}\right)^2\color{green}\sqrt{5}$$

90.$$\int_{0}^{\pi/2}\frac{\tan\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)}\ln^{n}(\cos x)\mathrm dx=2(-1)^n\color{green}\Gamma(n+1)\color{red}\eta(n)$$

91.$$\int_{0}^{\pi/2}\frac{\tan\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)}\ln(\sin x)\mathrm dx=\frac{1}{2}\color{green}-\ln 2$$

92.$$\int_{0}^{\pi/2}\frac{\tan\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)}\ln^{n}(\cos\left(\frac{x}{2}\right))\mathrm dx=\color{green}\frac{1}{2^{n-1}}\color{red}\left[\color{purple}(-1)^n\frac{n!}{2}+\color{blue}\sum_{j=1}^{n}\frac{n!}{j!}(-1)^{n+j}\ln^j(2)\right]$$

93.$$\int_{0}^{\pi/2}\ln(1+\alpha\cos x)\cdot \frac{\tan\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)}\mathrm dx=\ln\left[\left(\frac{\alpha+1}{4}\right)\cdot\left(\frac{\alpha+1}{2}\right)^{2/(\alpha-1)}\right]$$

94.$$\int_{0}^{\pi/2}\ln\left(1+\alpha\sin^2\left(\frac{x}{2}\right)\right)\cdot \frac{\tan\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)}\mathrm dx=\color{red}\ln\left[\left(\frac{\alpha+2}{4}\right)^{\color{green}\frac{\alpha+2}{\alpha+1}}\right]$$

95.$$\int_{0}^{\pi/2}\ln\left(1+\alpha\tan^2\left(\frac{x}{2}\right)\right)\cdot \frac{\tan\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)}\mathrm dx=\frac{1}{\alpha}\color{purple}\ln(\alpha+1)^{\alpha+1}-1$$

96.$$\int_{0}^{\pi/2}\ln\left(1+\alpha\tan\left(\frac{x}{2}\right)\right)\cdot \frac{\tan\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)}\mathrm dx=\frac{\alpha^2-1}{\alpha^2}\ln(\alpha+1)-\color{blue}\sum_{j=1}^{\alpha-2}\frac{1}{(j+1)(j+2)}$$ ,$$\alpha\ge3$$

97.$$\int_{0}^{\pi/2}\ln^2\left(1+\alpha\tan^2\left(\frac{x}{2}\right)\right)\cdot \frac{\tan\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)}\mathrm dx=\frac{\alpha+1}{\alpha}\ln^2(\alpha+1)-\frac{2}{\alpha}\color{purple}\ln(\alpha+1)^{\alpha+1}+2$$

98.$$\int_{0}^{\pi/2}\ln^3\left(1+\alpha\tan^2\left(\frac{x}{2}\right)\right)\cdot \frac{\tan\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)}\mathrm dx=\frac{\alpha+1}{\alpha}\ln^3(\alpha+1)-3\frac{\alpha+1}{\alpha}\ln^2(\alpha+1)+6\frac{\alpha+1}{\alpha}\ln(\alpha+1)-6$$

99.$$\int_{0}^{\pi/2}\ln^n\left(1+\alpha\tan^2\left(\frac{x}{2}\right)\right)\cdot \frac{\tan\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)}\mathrm dx=n!\left[(-1)^n+\color{blue}\frac{\alpha+1}{\alpha}\color{purple}\sum_{j=1}^{n}(-1)^{j+n}\frac{\ln^{j}(\alpha+1)}{j}\right]$$

100.$$\int_{0}^{\infty}\frac{e^{-x}\ln(1+e^{-x})}{\cosh x}\mathrm dx=\frac{\color{red}\pi^2+\color{blue}12\ln^2(2)}{48}$$

1.$$\int_{0}^{\infty}e^{-x}\frac{\cos\left(\frac{x}{\alpha}\right)-\cos(\alpha x)}{x}\mathrm dx=\color{blue}\ln\alpha$$

2.$$\int_{0}^{\infty}e^{-x}\frac{\cos(ax)-\cos(bx)}{x}\mathrm dx=\color{blue}\frac{1}{2}\color{green}\ln\frac{b^2+1}{a^2+1}$$

3.$$\int_{0}^{\infty}e^{-x}\frac{\cos^2(ax)-\cos^2(bx)}{x}\mathrm dx=\color{blue}\frac{1}{4}\color{purple}\ln\frac{4b^2+1}{4a^2+1}$$

4.$$\int_{0}^{\pi}\ln(3-2\cos x)\mathrm dx=\color{blue}2\pi\ln \color{red}\phi$$

5.$$\int_{0}^{\pi}\sin(x)\ln^n[a+(a-1)\cos x]\mathrm dx=2(-1)^n\color{red}n!+(-1)^{n-1}n!\frac{\alpha}{\beta}\color{blue}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k!}\ln^k(\alpha)$$

5.$$\alpha=2a-1$$ and $$\beta=a-1$$

6.$$\int_{0}^{1/{2\alpha}}\ln(1-\alpha x)\ln(1-2\alpha x)\mathrm dx=\frac{1}{24\color{red}\alpha}\color{green}\left(24-\pi^2-12\ln 2\right)$$

7.$$\int_{0}^{1/{2\alpha}}\ln(1-\alpha x)\ln(1-2\alpha x)\frac{\mathrm dx}{x^2}=\frac{\color{green}\alpha}{4}\color{blue}\left(\pi^2-4\ln^2 2\right)$$

8.$$\int_{0}^{\infty}\cos(x)\cos\left(\frac{x}{2}\right)\cdot\frac{x\cos(x)-\sin(x)}{x^3}\mathrm dx=-\frac{3}{2^5}\color{red}\pi$$

9.$$\int_{0}^{\infty}\cos(2x)\cos(x)\cos\left(\frac{x}{2}\right)\cdot\frac{x\cos(x)-\sin(x)}{x^3}\mathrm dx=-\frac{3}{2^6}\color{green}\pi$$

10.$$\int_{0}^{\infty}\cos(x)\cos(2x)\cdot\frac{x\cos^2(x)-\sin(x)}{x^3}\mathrm dx=\int_{0}^{\infty}\cos(x)\cos(2x)\cdot\frac{x\cos^3(x)-\sin(x)}{x^3}\mathrm dx=-\frac{1}{2^3}\color{green}\pi$$

11.$$\int_{0}^{\infty}\cos(x)\cos(2x)\cdot\frac{x\cos^4(x)-\sin(x)}{x^3}\mathrm dx=\int_{0}^{\infty}\cos(x)\cos(2x)\cdot\frac{x\cos^5(x)-\sin(x)}{x^3}\mathrm dx=-\frac{1}{2^2}\color{brown}\pi$$

12.$$\int_{0}^{\infty}\cos(x)\cos(2x)\cos(4x)\cdot\frac{x\cos^2(x)-\sin(x)}{x^3}\mathrm dx=-\frac{1}{2^4}\color{purple}\pi$$

13.$$\int_{0}^{\infty}\sin(x)\sin(2x)\sin(4x)\cdot\frac{x\cos^n(x)-\sin(x)}{x^2}\mathrm dx=\frac{1}{2^4}\color{purple}\pi$$

13. for n:={1,2,3 and 4}

14.$$\int_{0}^{\infty}\sin(x)\sin(2x)\sin(4x)\cdot\frac{x\sec^2(x)-\sin(x)}{x^2}\mathrm dx=\color{green}-\pi$$

15.$$\int_{0}^{\infty}\sin(x)\sin(2x)\sin(4x)\cdot\frac{x\cos^2(x)-\sin(x)}{x^3}\mathrm dx=-\frac{2\ln 2}{2}+\color{red}\frac{3\ln 3}{4}-\color{purple}\frac{5\ln 5}{8}+\color{blue}\frac{7\ln 7}{16}+\color{green}\ln\left(\frac{27\sqrt{27}}{2^7}\right)$$

16.$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{1-x^2}{1-y^2}\right)\frac{\mathrm dx \mathrm dy}{x^2-y^2}=-\color{red}\frac{7}{4}\color{grey}\zeta(3)$$

17.$$\int_{0}^{1}\int_{0}^{1}\ln\left(\frac{1-x}{1-y}\right)\frac{\mathrm dx \mathrm dy}{x-y}=-\color{red}\frac{8}{4}\color{brown}\zeta(2)$$

18.$$\int_{0}^{\pi}\sin(x)\sin\left(\frac{x}{2}\right)\ln^{2n-1}\left[\sin^2\left(\frac{x}{2}\right)\right]\mathrm dx=-\color{blue}\frac{(2n-1)!}{2^{2n-1}}$$

19.$$\int_{0}^{\pi}\sin(x)\sin\left(\frac{x}{2}\right)\ln^{2n}\left[\sin^2\left(\frac{x}{2}\right)\right]\mathrm dx=\color{purple}\frac{(2n)!}{2^{2n}}$$

20.$$\int_{0}^{\pi/2}\sin(x)\sin\left(\frac{x}{2}\right)\ln^2\left[\sin^2\left(\frac{x}{2}\right)\right]\mathrm dx=\color{blue}\frac{1}{8}\left[1+\color{red}2\ln 2+\color{green}2\ln^2(2)\right]$$

21.$$\int_{0}^{\pi}\sec(x)\sqrt{\tan\left(\frac{x}{2}\right)}\ln\tan\left(\frac{x}{2}\right)\mathrm dx=-\color{blue}\pi^2$$

22.$$\int_{0}^{\pi}\sec(x)\sqrt{\tan\left(\frac{x}{2}\right)}\ln^2\tan\left(\frac{x}{2}\right)\mathrm dx=-\color{green}\pi^3$$

23.$$\int_{0}^{\pi}\sec(x)\sqrt{\tan\left(\frac{x}{2}\right)}\ln^3\tan\left(\frac{x}{2}\right)\mathrm dx=-2\color{purple}\pi^4$$

24.$$\int_{0}^{\pi}\sec(x)\sqrt{\tan\left(\frac{x}{2}\right)}\ln^n\tan\left(\frac{x}{2}\right)\mathrm dx=-C_{n}\color{blue}\pi^{n+1}$$

24. Where $$C_{n}:=1,1,2,5,14,...$$ is the Catalan number

25.$$\int_{0}^{1}\frac{\operatorname{Li_{2}(1-x)}\ln^2(1-x)}{x^2}\mathrm dx=\color{green}\zeta(4)$$

26.$$\int_{0}^{\pi}\frac{\cos(nx)}{1+\sin(\alpha\beta)\cos(x)}\mathrm dx=-\color{red}\pi\color{blue}\tan^{n}\left(\frac{1}{2}\alpha\beta\right)\color{purple}\sec(\alpha\beta)$$

26.$$(\alpha,\beta,n):=1,2,3,...$$

27.$$\int_{0}^{1}\int_{0}^{1}\operatorname{Li_{n}(x)Li_{m}(y)}\frac{\mathrm dx \mathrm dy}{xy}=\zeta(n+1)\zeta(m+1)$$

28.$$\int_{0}^{1}\int_{0}^{1}\operatorname{Li_{2}(1-x)Li_{2}(1-y)}\frac{\mathrm dx \mathrm dy}{1-y}=\zeta(2)-\zeta(3)$$

29.$$\int_{0}^{\pi}[1-\cos(\alpha x)]^n\frac{\mathrm dx}{1-\cos x}=\color{blue}\frac{(2n-3)!!}{(2n-2)!!}\cdot 2^{n-1}\color{red}\alpha\color{green}\pi$$

30.$$\int_{0}^{\infty}e^{-x\pi}\cdot\frac{\sinh(3x\pi)}{\sinh(4x\pi)}\mathrm dx=\color{green}\frac{1}{2^4}\left(1!+\frac{3!}{\pi}\cdot\ln(2!)\right)$$

31.$$\int_{0}^{\infty}xe^{-3x\pi}\cdot\frac{\sinh(3x\pi)}{\sinh(4x\pi)}\mathrm dx=\color{green}\frac{2-C}{8\pi^2}-\frac{1}{128}$$

32.$$\int_{0}^{\infty}xe^{-5x\pi}\cdot\frac{\sinh(3x\pi)}{\sinh(4x\pi)}\mathrm dx=\color{green}\frac{1-2C}{16\pi^2}+\frac{1}{128}$$

33.$$\int_{0}^{\infty}\sin^2(\alpha x^2)\sin(2\alpha x^2)\frac{\mathrm dx}{x}=\color{purple}\frac{\pi}{16}$$

34.$$\int_{0}^{\infty}\sin^2(\alpha x^2)\sin(2\alpha x^2)\frac{\mathrm dx}{x^3}=\frac{\alpha}{2}\cdot\color{red}\ln 2$$

35.$$\int_{0}^{\infty}\sin^2(\alpha x^2)\sin(2\alpha x^2)\frac{\mathrm dx}{x^5}=\frac{\alpha^2}{4}\cdot\color{green}\pi$$

36.$$\int_{0}^{\infty}\sin^3( x^2)\sin(2x^2)\sin(4x^2)\frac{\mathrm dx}{x^5}=\frac{1}{64}\cdot\color{green}\pi$$

37.$$\int_{0}^{\pi}\cos x\ln(\alpha^2+2\alpha\cos x+1)\mathrm dx=\color{purple}\frac{\pi}{\alpha}$$

39.$$\int_{0}^{\infty}\sin(2^5x^2)\sin(2^7x^2)\sin(2^8x^2)\frac{\mathrm dx}{x^2}=\sqrt{\pi}(-\color{blue}\sqrt{3}+\color{green}\sqrt{5}+\color{purple}\sqrt{7}-\color{red}\sqrt{13})$$

40.$$\int_{0}^{1}(1-x^{2^{n}})^{(2^n\alpha-1)/2^n}\mathrm dx=\color{red}\pi\csc\left(\frac{\pi}{2^n}\right)\color{blue}\cdot\frac{(2^n\alpha-1)!!}{(2n)!!}\color{green}\cdot\frac{1}{2^{{(n-1)\alpha}+n}}$$

41.$$\int_{0}^{\pi/2}\ln^n\left[\tan\left(\frac{x}{2}\right)\right]\frac{\mathrm dx}{\cos^2\left(\frac{x}{2}\right)}=2(-1)^n\color{red}n!$$

42.$$\int_{0}^{\pi/2}\ln^2\left[\tan\left(\frac{x}{2}\right)\right]\frac{\mathrm dx}{\cos^2(x)}=\color{red}\frac{\pi^2}{2}$$

43.$$\int_{0}^{\pi/2}\ln^4\left[\tan\left(\frac{x}{2}\right)\right]\frac{\mathrm dx}{\cos^2(x)}=\color{green}\frac{\pi^4}{2}$$

44.$$\int_{0}^{\pi/2}\ln^{2n}\left[\tan\left(\frac{x}{2}\right)\right]\frac{\mathrm dx}{\cos^2(x)}=\color{green}\pi^{2n}\color{purple}F(n)$$

45.$$\int_{0}^{\pi/2}\arctan(a\tan^2 x)\frac{\mathrm dx}{b\sin^2 x+c\cos^2 x}=\frac{\pi}{\sqrt{bc}}\left(\arctan\left(1+\sqrt{\frac{2ac}{b}}\right)-\frac{\pi}{4}\right)$$

46.$$\int_{0}^{\infty}\left(\frac{\arctan x}{1+(x+\frac{1}{x})\arctan x}\right)^2=\color{blue}\frac{\pi}{4}-\color{red}\frac{1}{\pi}$$

47.$$\int_{0}^{\infty}\left(\frac{\arctan x}{1+x\arctan x}\right)^2=\color{brown}\frac{2}{\pi}$$

48.$$\int_{0}^{1}\left(\frac{\arctan x}{1+x\arctan x}\right)^2=\color{brown}\frac{\color{green}{4-\pi}}{\color{blue}{4+\pi}}$$

49.$$\int_{0}^{\pi/2}\ln(\sin^2 x+\alpha^2\cos^2 x)\frac{\mathrm dx}{\cos^2 x}=\color{green}(\alpha-1)\color{blue}\pi$$

50.$$\int_{0}^{\pi/2}\ln[1+4\alpha(\alpha-1)\cos^2 x]\frac{\mathrm dx}{\cos^2 x}=2\color{green}(\alpha-1)\color{blue}\pi$$

51.$$\int_{0}^{\pi}\left(\frac{1-\frac{1}{\alpha}\cos x}{1+\frac{1}{\alpha^2}\cos^2 x}\right)^2=\color{red}\frac{\alpha}{\sqrt{1+\alpha^2}}\color{blue}\pi$$

52.$$\int_{0}^{\pi}\sin x\left(\frac{1-\frac{1}{\alpha}\cos x}{1+\frac{1}{\alpha^2}\cos^2 x}\right)^2=2\color{red}\alpha\arccot(\alpha)$$

53.$$\int_{0}^{\pi}\cos x\ln\left(\frac{1-\frac{1}{\alpha}\cos x}{1+\frac{1}{\alpha^2}\cos^2 x}\right)=\color{blue}(\color{red}{\sqrt{\alpha^2-1}-\alpha)}\color{green}\pi$$

54.$$\int_{0}^{\pi}\sin x\ln\left(\frac{1-\cos x}{1+\cos^2 x}\right)\mathrm dx=\color{red}2-\color{blue}\pi$$

55.$$\int_{0}^{\pi/2}\ln^2\left[\tan^2\left(\frac{x}{2}\right)\right]\mathrm dx=\color{red}\frac{\pi^3}{2}$$

56.$$\int_{0}^{1}\frac{1}{x\sqrt{1-x}}\ln^n\left(\frac{1+\sqrt{1-x}}{2\sqrt{1-x}}\right)\mathrm dx=\color{purple}\Gamma(n+1)\zeta(n+1)$$

57.$$\int_{0}^{1}\frac{1}{\sqrt{1-x}}\ln\left(\frac{1+\sqrt{1-x}}{2\sqrt{1-x}}\right)\mathrm dx=\color{brown}\ln 4$$

58.$$\int_{0}^{1}\frac{1}{\sqrt{1-x}}\ln^2\left(\frac{1+\sqrt{1-x}}{2\sqrt{1-x}}\right)\mathrm dx=\color{red}2\zeta(2)-\color{orange}2\ln^2(2)$$

59.$$\int_{0}^{1}\frac{1}{\sqrt{1-x}}\ln\left(\frac{1-\sqrt{1-x}}{2\sqrt{1-x}}\right)\mathrm dx=-\color{red}\ln 4$$

60.$$\int_{0}^{1}\frac{1}{\sqrt{1-x}}\ln^2\left(\frac{1-\sqrt{1-x}}{2\sqrt{1-x}}\right)\mathrm dx=\color{blue}4\zeta(2)+\color{green}2\ln^2(2)$$

61.$$\int_{0}^{1}\frac{\ln(1-x)}{\sqrt{1-x}}\ln\left(\frac{1-\sqrt{1-x}}{2\sqrt{1-x}}\right)\mathrm dx=-\color{blue}2\zeta(2)-\color{grey}4\ln(2)$$

62.$$\int_{0}^{-1}\frac{e^{x}+e^{1/x}-1}{x}\cdot \ln^2(-x)\mathrm dx=\color{green}2\cdot\frac{\zeta(3)}{3}+\color{blue}\frac{\gamma^3}{3}+\color{red}\gamma\zeta(2)$$

63.$$\int_{0}^{-1}\frac{e^{x}+xe^{1/x}-1}{x}\mathrm dx=\color{green}\gamma+\color{blue}\frac{1}{e}$$

64.$$\int_{0}^{-1}\frac{e^{2x}+\frac{1}{2}xe^{2/x}-1}{x}\mathrm dx=\color{red}\gamma+\color{brown}\frac{1}{2e^2}+\color{grey}\ln 2$$

65.$$\int_{0}^{-1}\frac{e^{\alpha x}+\frac{1}{\alpha}xe^{\alpha/x}-1}{x}\mathrm dx=\color{red}\gamma+\color{brown}\frac{1}{\alpha e^{\alpha}}+\color{grey}\ln(\alpha)$$

66.$$\int_{0}^{\infty}\frac{\ln(1-e^{-\alpha x})}{1-e^{\alpha x}}x\mathrm dx=\color{green}\frac{1}{\alpha^2}\zeta(3)$$

67.$$\int_{0}^{\infty}\frac{\ln(1-e^{-\alpha x})}{1-e^{\alpha x}}x^2\mathrm dx=\color{blue}\frac{1}{2\alpha^3}\zeta(4)$$

68.$$\int_{0}^{\infty}\frac{\ln(1-e^{-\alpha x})}{1-e^{-\alpha x}}x\mathrm dx=-\color{green}\frac{2}{\alpha^2}\zeta(3)$$

69.$$\int_{0}^{\infty}\frac{\ln(1-e^{-\alpha x})}{1-e^{-\alpha x}}x^2\mathrm dx=-\color{red}\frac{5}{2\alpha^3}\zeta(4)$$

70.$$\int_{0}^{1}\ln^{n}(x)\arctan\left(x^m\right)\frac{\mathrm dx}{x}=-\color{red}\frac{1}{m^{n+1}}\color{green}\Gamma(n+1)\color{purple}\beta(n+2)$$

71.$$\int_{0}^{\infty}e^{-x}\ln(1+\cosh x)\mathrm dx=\color{red}\ln 8 -1$$

72.$$\int_{0}^{\infty}e^{-x}\ln(1+\operatorname{sech x})\mathrm dx=\color{red}\ln 8 -\frac{\pi}{2}$$

73.$$\int_{0}^{\infty}e^{-x}\ln^2(1+\operatorname{sech x})\mathrm dx=\color{green}2\zeta(2)-4C+\pi\ln(2)+\ln(2)\ln(4)$$

74.$$\int_{0}^{\infty}e^{-x}\ln(1-\operatorname{sech x})\mathrm dx=\color{purple}-\ln 2 -\frac{\pi}{2}$$

75.$$\int_{0}^{\infty}e^{-x}\ln\left(\frac{1+\operatorname{sech x}}{1-\operatorname{sech x}}\right)\mathrm dx=\color{red}\ln(16)$$

76.$$\int_{0}^{1}\frac{2}{(2-x)^2}\ln x\mathrm dx=\color{green}\ln 2$$

77.$$\int_{0}^{1}\frac{2^3}{(2-x)^3}\ln x\mathrm dx=1+\color{green}\ln 2$$

78.$$\int_{0}^{\pi}\frac{\mathrm dx}{\alpha+\alpha\sin x+\cos x}=\color{purple}\ln\left(\frac{\alpha+1}{\alpha-1}\right)$$

79.$$\int_{0}^{\pi}\frac{\mathrm dx}{3+2\sin x+\cos x}=\color{purple}\frac{\pi}{4}$$

80.$$\int_{0}^{\pi}\frac{\mathrm dx}{\color{red}{5}+\color{blue}{4}\sin x+\color{green}{3}\cos x}=\color{red}\frac{1}{2!}$$

81.$$\int_{0}^{\pi/2}\frac{\mathrm dx}{\color{red}{5}+\color{blue}{4}\sin x+\color{green}{3}\cos x}=\color{red}\frac{1}{3!}$$

82.$$\int_{0}^{\infty}\sin(x)\sin(\alpha x)\ln(x)\frac{\mathrm dx}{x^2}=\color{red}\frac{\pi}{4}\left(2(1-\gamma)+\color{blue}\ln\left[\frac{(\alpha-1)^{\alpha-1}}{(\alpha+1)^{\alpha+1}}\right]\right)$$

83.$$\int_{0}^{\pi}\ln(1+\sin\alpha\cos x)\frac{\mathrm dx}{\cos x}=\color{red}\pi(\pi-\alpha)$$

84.$$\int_{0}^{\pi/2}\ln(1+\sin\alpha\cos x)\frac{\mathrm dx}{\cos x}=\color{red}(\pi-\alpha)\cdot\color{green}\frac{\alpha}{2}$$

85.$$\int_{0}^{\pi/2}\ln(1+\cos\alpha\cos x)\frac{\mathrm dx}{\cos x}=\frac{1}{2}\color{red}\left(\frac{\pi}{2}-\alpha\right)\color{blue}\left(\frac{\pi}{2}+\alpha\right)$$

86.$$\int_{0}^{\infty}\frac{1-\cos(\alpha x)}{x^{1+\beta}}\mathrm dx=\color{purple}\frac{\alpha^{\beta}}{\beta}\color{green}\cos\left(\frac{\pi \beta}{2}\right)\color{red}\Gamma(1-\beta)$$

86.$$0<\beta<1$$

87.$$\int_{0}^{\infty}\frac{1-\cos(\alpha x)}{x^{1+\beta}}\mathrm dx=-\color{purple}\frac{\alpha^{\beta}}{\beta(\beta-1)}\color{green}\cos\left(\frac{\pi \beta}{2}\right)\color{red}\Gamma(2-\beta)$$

87.$$0<\beta<2$$

88.$$\int_{0}^{\pi}\frac{\mathrm dx}{(\alpha+\sin^2 x)^2}=\color{red}\frac{\pi}{2}\cdot\color{green}\frac{2\alpha+1}{\sqrt[3]{(a^2+a)^2}}$$

89.$$\int_{0}^{\infty}\frac{\ln x}{(1-x^2)(1+x^2)^2}\mathrm dx=-\color{red}\frac{\pi}{16}(\pi+2)$$

90.$$\int_{0}^{\infty}\frac{\ln x}{(1-x^2)(1+x)^2}\mathrm dx=-\color{blue}\frac{1}{16}(\pi^2+2^2)$$

91.$$\int_{0}^{\infty}\frac{\ln x}{(1-x^2)(1+x^{\alpha})}\mathrm dx=\color{purple}\frac{\pi^2}{8}$$

92.$$\int_{0}^{\infty}\frac{x^2\ln x}{(1-x)^{\color{purple}1}(1+x^2)^{\color{red}2}(1+x)^{\color{blue}3}}\mathrm dx=-\color{purple}\frac{1}{16}$$

93.$$\int_{\pi/2}^{\pi/2}\cos(\alpha+\beta\tan x)\mathrm dx=\color{green}\frac{\pi}{e^{\beta}}\color{blue}\cos(\alpha)$$

94.$$\int_{\pi/2}^{\pi/2}\sin(\alpha+\beta\tan x)\mathrm dx=\color{black}\frac{\pi}{e^{\beta}}\color{green}\sin(\alpha)$$

95.$$\int_{\pi/2}^{\pi/2}\cos(\tan(4x))\mathrm dx=\color{green}\frac{\pi}{2e}$$

96.$$\int_{\pi/2}^{\pi/2}\cos(\tan(16x))\mathrm dx=\color{brown}\frac{\pi}{8e}$$

97.$$\int_{\pi/2}^{\pi/2}\cos^2(\tan x)\mathrm dx=\color{blue}\frac{\pi}{e}\color{red}\cosh(1)$$

98.$$\int_{\pi/2}^{\pi/2}\sin^2(\tan x)\mathrm dx=\color{orange}\frac{\pi}{e}\color{grey}\sinh(1)$$

99.$$\int_{\pi/2}^{\pi/2}\cos^2 x\cos(\alpha+\beta\tan x)\mathrm dx=\frac{1+\beta}{2}\cdot\color{purple}\frac{\pi}{e^{\beta}}\color{red}\cos(\alpha)$$

100.$$\int_{\pi/2}^{\pi/2}\sin^2 x\cos(\alpha+\beta\tan x)\mathrm dx=\frac{1-\beta}{2}\cdot\color{purple}\frac{\pi}{e^{\beta}}\color{red}\cos(\alpha)$$

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